Mixing
Link variables between two text files
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
The following explanation is just a representation for what I would like to achieve.
I have two text files:
The first text file log1.txt contain the following entries:
Black
Blue
Brown
Copper
Cyan
Gold
Gray
Green
The second text file log2.txt contain the following entries:
Ugly
Nice
cool
pretty
I would like to read both texts at the same time and generate the following output:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
How can I achieve the previous output using bash or shell? I tried to apply two loops at the same time :for loop" and/orwhile loop` but did not work! For example, I tried this awkward code (I am newbie in coding):
#!/bin/bash
while IFS= read -r line; do
for ii in $(cat log1.txt); do
echo "The first color "$i "is" $ii
done <log2.txt
done
I have no idea or clue how to change between "first color", "second color", ….etc
bash shell-script
asked Sep 7 at 13:13
Kasper
12611
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Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
5
down vote
favorite
The following explanation is just a representation for what I would like to achieve.
I have two text files:
The first text file log1.txt contain the following entries:
Black
Blue
Brown
Copper
Cyan
Gold
Gray
Green
The second text file log2.txt contain the following entries:
Ugly
Nice
cool
pretty
I would like to read both texts at the same time and generate the following output:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
How can I achieve the previous output using bash or shell? I tried to apply two loops at the same time :for loop" and/orwhile loop` but did not work! For example, I tried this awkward code (I am newbie in coding):
#!/bin/bash
while IFS= read -r line; do
for ii in $(cat log1.txt); do
echo "The first color "$i "is" $ii
done <log2.txt
done
I have no idea or clue how to change between "first color", "second color", ….etc
bash shell-script
asked Sep 7 at 13:13
Kasper
12611
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
2
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
1
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
The following explanation is just a representation for what I would like to achieve.
I have two text files:
The first text file log1.txt contain the following entries:
Black
Blue
Brown
Copper
Cyan
Gold
Gray
Green
The second text file log2.txt contain the following entries:
Ugly
Nice
cool
pretty
I would like to read both texts at the same time and generate the following output:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
How can I achieve the previous output using bash or shell? I tried to apply two loops at the same time :for loop" and/orwhile loop` but did not work! For example, I tried this awkward code (I am newbie in coding):
#!/bin/bash
while IFS= read -r line; do
for ii in $(cat log1.txt); do
echo "The first color "$i "is" $ii
done <log2.txt
done
I have no idea or clue how to change between "first color", "second color", ….etc
bash shell-script
asked Sep 7 at 13:13
Kasper
12611
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The following explanation is just a representation for what I would like to achieve.
I have two text files:
The first text file log1.txt contain the following entries:
Black
Blue
Brown
Copper
Cyan
Gold
Gray
Green
The second text file log2.txt contain the following entries:
Ugly
Nice
cool
pretty
I would like to read both texts at the same time and generate the following output:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
How can I achieve the previous output using bash or shell? I tried to apply two loops at the same time :for loop" and/orwhile loop` but did not work! For example, I tried this awkward code (I am newbie in coding):
#!/bin/bash
while IFS= read -r line; do
for ii in $(cat log1.txt); do
echo "The first color "$i "is" $ii
done <log2.txt
done
I have no idea or clue how to change between "first color", "second color", ….etc
bash shell-script
asked Sep 7 at 13:13
Kasper
12611
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Sep 7 at 14:51
asked Sep 7 at 13:13
Kasper
12611
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 7 at 13:13
Kasper
12611
asked Sep 7 at 13:13
Kasper
12611
12611
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kasper is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
2
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
1
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22
add a comment |Â
Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
2
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
1
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22
Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
2
2
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
1
1
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
With some modifications on @Sam's code, you can achieve what you want using case control structure as follows:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
# Case control structure to replace digit equivalent in words
case $i in
1) echo -n "The first color ";;
2) echo -n "The second color ";;
3) echo -n "The third color ";;
4) echo -n "The fourth color ";;
5) echo -n "The fifth color ";;
6) echo -n "The sixth color ";;
7) echo -n "The seventh color ";;
8) echo -n "The eighth color ";;
9) echo -n "The ninth color ";;
10) echo -n "The tenth color ";;
11) echo -n "The eleventh color ";;
esac
echo $i"$i$arg1 is $arg2" | tr -d '0123456789'
done
The output is as follows:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
edited Sep 7 at 18:15
wjandrea
462312
answered Sep 7 at 14:19
Goro
1,34541643
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why thedigit=$(echo $i)? andecho $digit | tr -d '0123456789'?
– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if youecho $digitthe output is something likeThe first color 1. Given that the OP would like to seeThe first color black is uglyand to get around that just trim the numbers out
– Goro
Sep 7 at 15:00
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also whydigit=$(echo $i)and notdigit=$i(orfor digitinstead offor i).
– Stéphane Chazelas
Sep 7 at 15:04
 |Â
show 1 more comment
up vote
4
down vote
Your shell does not know English, so automatically generating the spelled-out numbers with correct suffixes to an arbitrary count would involve some additional effort. With just digits for the numbering and the additional presumption that log1.txt is the longer file, try this:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
echo "Color No. $i $arg1 is $arg2."
done
answered Sep 7 at 13:56
Sam
1436
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
add a comment |Â
up vote
4
down vote
With zsh and with libnumbertext-tools's spellout on Debian:
#! /bin/zsh -
colors=($(f)"$(<log1.txt)")
adjectives=($(f)"$(head -n $#colors <log2.txt)")
/usr/lib/libnumbertext/spellout -l /usr/share/libnumbertext/en
-p ordinal 1-$#colors |
for color adjective in $colors:^^adjectives; do
read num &&
print -r The $num color $color is $adjective
done
(note that it's US-English. For instance, for 101, you get one hundred first instead of one hundred and first)
If you can't install zsh or any software that spells out numbers but have a list of English ordinals in a third log3.txt file, you could do in most shells including bash (Bourne-like, rc-like, fish at least):
#! /bin/sh -
awk '
BEGIN while ((getline a < "log2.txt") > 0) adjective[na++] = a
if ((getline num < "log3.txt") <= 0) num = NR "th"
print "The "num" color "$0" is "adjective[(NR-1)%na]
' log1.txt
(falling back to <digits>th if we run out of English numbers).
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
@Kasper, trysudo yum install zsh, or install in your home directory from source. You'd need thatspelloututility as well.
– Stéphane Chazelas
Sep 7 at 13:54
No creative use ofpasteand a multiply-repeatedlog2.txt? (-:
– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
 |Â
show 1 more comment
up vote
-1
down vote
I am surprised nobody has suggested using arrays. Here is my crude attempt (using the log3.txt idea from @Stephane above.
#!/bin/bash
nl1=$( wc -l < log1.txt )
nl2=$( wc -l < log2.txt )
nlnums=$( wc -l < nums.txt )
declare -a arr1[$nl1]
declare -a arr2[$nl2]
declare -a nums[$nlnums]
for (( i=0; i < $nl1; i++ ))
do
read arr1[$i]
done < log1.txt
for (( i=0; i < $nl2; i++ ))
do
read arr2[$i]
done < log2.txt
for (( i=0; i < $nlnums; i++ ))
do
read nums[$i]
done < nums.txt
j=0
for (( i=0; i < $nl1; i++ ))
do
echo "The $nums[$i] color $arr1[$i] is $arr2[$j]"
j=$(( (j+1) % $nl2 ))
done
The file nums.txt is as follows:
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
The code needs to be cleaned a bit but illustrates the point.
answered Sep 7 at 22:32
unxnut
3,3802918
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With some modifications on @Sam's code, you can achieve what you want using case control structure as follows:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
# Case control structure to replace digit equivalent in words
case $i in
1) echo -n "The first color ";;
2) echo -n "The second color ";;
3) echo -n "The third color ";;
4) echo -n "The fourth color ";;
5) echo -n "The fifth color ";;
6) echo -n "The sixth color ";;
7) echo -n "The seventh color ";;
8) echo -n "The eighth color ";;
9) echo -n "The ninth color ";;
10) echo -n "The tenth color ";;
11) echo -n "The eleventh color ";;
esac
echo $i"$i$arg1 is $arg2" | tr -d '0123456789'
done
The output is as follows:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
edited Sep 7 at 18:15
wjandrea
462312
answered Sep 7 at 14:19
Goro
1,34541643
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why thedigit=$(echo $i)? andecho $digit | tr -d '0123456789'?
– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if youecho $digitthe output is something likeThe first color 1. Given that the OP would like to seeThe first color black is uglyand to get around that just trim the numbers out
– Goro
Sep 7 at 15:00
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also whydigit=$(echo $i)and notdigit=$i(orfor digitinstead offor i).
– Stéphane Chazelas
Sep 7 at 15:04
 |Â
show 1 more comment
up vote
4
down vote
accepted
With some modifications on @Sam's code, you can achieve what you want using case control structure as follows:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
# Case control structure to replace digit equivalent in words
case $i in
1) echo -n "The first color ";;
2) echo -n "The second color ";;
3) echo -n "The third color ";;
4) echo -n "The fourth color ";;
5) echo -n "The fifth color ";;
6) echo -n "The sixth color ";;
7) echo -n "The seventh color ";;
8) echo -n "The eighth color ";;
9) echo -n "The ninth color ";;
10) echo -n "The tenth color ";;
11) echo -n "The eleventh color ";;
esac
echo $i"$i$arg1 is $arg2" | tr -d '0123456789'
done
The output is as follows:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
edited Sep 7 at 18:15
wjandrea
462312
answered Sep 7 at 14:19
Goro
1,34541643
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why thedigit=$(echo $i)? andecho $digit | tr -d '0123456789'?
– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if youecho $digitthe output is something likeThe first color 1. Given that the OP would like to seeThe first color black is uglyand to get around that just trim the numbers out
– Goro
Sep 7 at 15:00
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also whydigit=$(echo $i)and notdigit=$i(orfor digitinstead offor i).
– Stéphane Chazelas
Sep 7 at 15:04
 |Â
show 1 more comment
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With some modifications on @Sam's code, you can achieve what you want using case control structure as follows:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
# Case control structure to replace digit equivalent in words
case $i in
1) echo -n "The first color ";;
2) echo -n "The second color ";;
3) echo -n "The third color ";;
4) echo -n "The fourth color ";;
5) echo -n "The fifth color ";;
6) echo -n "The sixth color ";;
7) echo -n "The seventh color ";;
8) echo -n "The eighth color ";;
9) echo -n "The ninth color ";;
10) echo -n "The tenth color ";;
11) echo -n "The eleventh color ";;
esac
echo $i"$i$arg1 is $arg2" | tr -d '0123456789'
done
The output is as follows:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
edited Sep 7 at 18:15
wjandrea
462312
answered Sep 7 at 14:19
Goro
1,34541643
With some modifications on @Sam's code, you can achieve what you want using case control structure as follows:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
# Case control structure to replace digit equivalent in words
case $i in
1) echo -n "The first color ";;
2) echo -n "The second color ";;
3) echo -n "The third color ";;
4) echo -n "The fourth color ";;
5) echo -n "The fifth color ";;
6) echo -n "The sixth color ";;
7) echo -n "The seventh color ";;
8) echo -n "The eighth color ";;
9) echo -n "The ninth color ";;
10) echo -n "The tenth color ";;
11) echo -n "The eleventh color ";;
esac
echo $i"$i$arg1 is $arg2" | tr -d '0123456789'
done
The output is as follows:
The first color Black is Ugly
The second color Blue is Nice
The third color Brown is cool
The fourth color Copper is pretty
The fifth color Cyan is Ugly
The sixth color Gold is Nice
The seventh color Gray is cool
The eighth color Green is pretty
edited Sep 7 at 18:15
wjandrea
462312
answered Sep 7 at 14:19
Goro
1,34541643
edited Sep 7 at 18:15
wjandrea
462312
edited Sep 7 at 18:15
wjandrea
462312
edited Sep 7 at 18:15
wjandrea
462312
462312
answered Sep 7 at 14:19
Goro
1,34541643
answered Sep 7 at 14:19
Goro
1,34541643
answered Sep 7 at 14:19
Goro
1,34541643
1,34541643
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why thedigit=$(echo $i)? andecho $digit | tr -d '0123456789'?
– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if youecho $digitthe output is something likeThe first color 1. Given that the OP would like to seeThe first color black is uglyand to get around that just trim the numbers out
– Goro
Sep 7 at 15:00
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also whydigit=$(echo $i)and notdigit=$i(orfor digitinstead offor i).
– Stéphane Chazelas
Sep 7 at 15:04
 |Â
show 1 more comment
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why thedigit=$(echo $i)? andecho $digit | tr -d '0123456789'?
– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if youecho $digitthe output is something likeThe first color 1. Given that the OP would like to seeThe first color black is uglyand to get around that just trim the numbers out
– Goro
Sep 7 at 15:00
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also whydigit=$(echo $i)and notdigit=$i(orfor digitinstead offor i).
– Stéphane Chazelas
Sep 7 at 15:04
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why echo the digits and then delete them again?
– Sam
Sep 7 at 14:58
Why the
digit=$(echo $i)? and echo $digit | tr -d '0123456789'?– Stéphane Chazelas
Sep 7 at 14:59
Why the
digit=$(echo $i)? and echo $digit | tr -d '0123456789'?– Stéphane Chazelas
Sep 7 at 14:59
@Sam, if you
echo $digit the output is something like The first color 1 . Given that the OP would like to see The first color black is ugly and to get around that just trim the numbers out– Goro
Sep 7 at 15:00
@Sam, if you
echo $digit the output is something like The first color 1 . Given that the OP would like to see The first color black is ugly and to get around that just trim the numbers out– Goro
Sep 7 at 15:00
1
1
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
Note that it runs 8 commands for each line of log1. Using those head+tail for each line means the files have to be read from the start at each iteration which is going to be even less efficient than using while read loops
– Stéphane Chazelas
Sep 7 at 15:01
1
1
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also why
digit=$(echo $i) and not digit=$i (or for digit instead of for i).– Stéphane Chazelas
Sep 7 at 15:04
So why outputting that digit in the first place (twice with $digit$i) just to remove it later with tr (which would also remove digits in color names or adjectives). Also why
digit=$(echo $i) and not digit=$i (or for digit instead of for i).– Stéphane Chazelas
Sep 7 at 15:04
 |Â
show 1 more comment
up vote
4
down vote
Your shell does not know English, so automatically generating the spelled-out numbers with correct suffixes to an arbitrary count would involve some additional effort. With just digits for the numbering and the additional presumption that log1.txt is the longer file, try this:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
echo "Color No. $i $arg1 is $arg2."
done
answered Sep 7 at 13:56
Sam
1436
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
add a comment |Â
up vote
4
down vote
Your shell does not know English, so automatically generating the spelled-out numbers with correct suffixes to an arbitrary count would involve some additional effort. With just digits for the numbering and the additional presumption that log1.txt is the longer file, try this:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
echo "Color No. $i $arg1 is $arg2."
done
answered Sep 7 at 13:56
Sam
1436
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Your shell does not know English, so automatically generating the spelled-out numbers with correct suffixes to an arbitrary count would involve some additional effort. With just digits for the numbering and the additional presumption that log1.txt is the longer file, try this:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
echo "Color No. $i $arg1 is $arg2."
done
answered Sep 7 at 13:56
Sam
1436
Your shell does not know English, so automatically generating the spelled-out numbers with correct suffixes to an arbitrary count would involve some additional effort. With just digits for the numbering and the additional presumption that log1.txt is the longer file, try this:
#!/bin/bash
log1_length=$(wc -l <log1.txt)
log2_length=$(wc -l <log2.txt)
for i in $(seq $log1_length); do
arg1="$(head -$i <log1.txt | tail -1)"
arg2="$(head -$(((i-1) % log2_length + 1)) <log2.txt | tail -1)"
echo "Color No. $i $arg1 is $arg2."
done
answered Sep 7 at 13:56
Sam
1436
answered Sep 7 at 13:56
Sam
1436
answered Sep 7 at 13:56
Sam
1436
answered Sep 7 at 13:56
Sam
1436
1436
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
add a comment |Â
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
2
2
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
I used your variables names. Thank you ;-)
– Goro
Sep 7 at 14:24
Thank you very much!
– Kasper
Sep 8 at 0:13
Thank you very much!
– Kasper
Sep 8 at 0:13
add a comment |Â
up vote
4
down vote
With zsh and with libnumbertext-tools's spellout on Debian:
#! /bin/zsh -
colors=($(f)"$(<log1.txt)")
adjectives=($(f)"$(head -n $#colors <log2.txt)")
/usr/lib/libnumbertext/spellout -l /usr/share/libnumbertext/en
-p ordinal 1-$#colors |
for color adjective in $colors:^^adjectives; do
read num &&
print -r The $num color $color is $adjective
done
(note that it's US-English. For instance, for 101, you get one hundred first instead of one hundred and first)
If you can't install zsh or any software that spells out numbers but have a list of English ordinals in a third log3.txt file, you could do in most shells including bash (Bourne-like, rc-like, fish at least):
#! /bin/sh -
awk '
BEGIN while ((getline a < "log2.txt") > 0) adjective[na++] = a
if ((getline num < "log3.txt") <= 0) num = NR "th"
print "The "num" color "$0" is "adjective[(NR-1)%na]
' log1.txt
(falling back to <digits>th if we run out of English numbers).
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
@Kasper, trysudo yum install zsh, or install in your home directory from source. You'd need thatspelloututility as well.
– Stéphane Chazelas
Sep 7 at 13:54
No creative use ofpasteand a multiply-repeatedlog2.txt? (-:
– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
 |Â
show 1 more comment
up vote
4
down vote
With zsh and with libnumbertext-tools's spellout on Debian:
#! /bin/zsh -
colors=($(f)"$(<log1.txt)")
adjectives=($(f)"$(head -n $#colors <log2.txt)")
/usr/lib/libnumbertext/spellout -l /usr/share/libnumbertext/en
-p ordinal 1-$#colors |
for color adjective in $colors:^^adjectives; do
read num &&
print -r The $num color $color is $adjective
done
(note that it's US-English. For instance, for 101, you get one hundred first instead of one hundred and first)
If you can't install zsh or any software that spells out numbers but have a list of English ordinals in a third log3.txt file, you could do in most shells including bash (Bourne-like, rc-like, fish at least):
#! /bin/sh -
awk '
BEGIN while ((getline a < "log2.txt") > 0) adjective[na++] = a
if ((getline num < "log3.txt") <= 0) num = NR "th"
print "The "num" color "$0" is "adjective[(NR-1)%na]
' log1.txt
(falling back to <digits>th if we run out of English numbers).
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
@Kasper, trysudo yum install zsh, or install in your home directory from source. You'd need thatspelloututility as well.
– Stéphane Chazelas
Sep 7 at 13:54
No creative use ofpasteand a multiply-repeatedlog2.txt? (-:
– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
 |Â
show 1 more comment
up vote
4
down vote
up vote
4
down vote
With zsh and with libnumbertext-tools's spellout on Debian:
#! /bin/zsh -
colors=($(f)"$(<log1.txt)")
adjectives=($(f)"$(head -n $#colors <log2.txt)")
/usr/lib/libnumbertext/spellout -l /usr/share/libnumbertext/en
-p ordinal 1-$#colors |
for color adjective in $colors:^^adjectives; do
read num &&
print -r The $num color $color is $adjective
done
(note that it's US-English. For instance, for 101, you get one hundred first instead of one hundred and first)
If you can't install zsh or any software that spells out numbers but have a list of English ordinals in a third log3.txt file, you could do in most shells including bash (Bourne-like, rc-like, fish at least):
#! /bin/sh -
awk '
BEGIN while ((getline a < "log2.txt") > 0) adjective[na++] = a
if ((getline num < "log3.txt") <= 0) num = NR "th"
print "The "num" color "$0" is "adjective[(NR-1)%na]
' log1.txt
(falling back to <digits>th if we run out of English numbers).
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
With zsh and with libnumbertext-tools's spellout on Debian:
#! /bin/zsh -
colors=($(f)"$(<log1.txt)")
adjectives=($(f)"$(head -n $#colors <log2.txt)")
/usr/lib/libnumbertext/spellout -l /usr/share/libnumbertext/en
-p ordinal 1-$#colors |
for color adjective in $colors:^^adjectives; do
read num &&
print -r The $num color $color is $adjective
done
(note that it's US-English. For instance, for 101, you get one hundred first instead of one hundred and first)
If you can't install zsh or any software that spells out numbers but have a list of English ordinals in a third log3.txt file, you could do in most shells including bash (Bourne-like, rc-like, fish at least):
#! /bin/sh -
awk '
BEGIN while ((getline a < "log2.txt") > 0) adjective[na++] = a
if ((getline num < "log3.txt") <= 0) num = NR "th"
print "The "num" color "$0" is "adjective[(NR-1)%na]
' log1.txt
(falling back to <digits>th if we run out of English numbers).
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
edited Sep 7 at 16:07
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
answered Sep 7 at 13:42
Stéphane Chazelas
283k53521858
283k53521858
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
@Kasper, trysudo yum install zsh, or install in your home directory from source. You'd need thatspelloututility as well.
– Stéphane Chazelas
Sep 7 at 13:54
No creative use ofpasteand a multiply-repeatedlog2.txt? (-:
– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
 |Â
show 1 more comment
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
@Kasper, trysudo yum install zsh, or install in your home directory from source. You'd need thatspelloututility as well.
– Stéphane Chazelas
Sep 7 at 13:54
No creative use ofpasteand a multiply-repeatedlog2.txt? (-:
– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
unfortunately I don't have zsh on Centos 7 to test it. thank you
– Kasper
Sep 7 at 13:52
1
1
@Kasper, try
sudo yum install zsh, or install in your home directory from source. You'd need that spellout utility as well.– Stéphane Chazelas
Sep 7 at 13:54
@Kasper, try
sudo yum install zsh, or install in your home directory from source. You'd need that spellout utility as well.– Stéphane Chazelas
Sep 7 at 13:54
No creative use of
paste and a multiply-repeated log2.txt? (-:– JdeBP
Sep 7 at 14:11
No creative use of
paste and a multiply-repeated log2.txt? (-:– JdeBP
Sep 7 at 14:11
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
@JdeBP, the multiply-repeated would get ugly anyway.
– Stéphane Chazelas
Sep 7 at 14:15
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
Thank you @StéphaneChazelas will try it for sure!
– TNT
Sep 7 at 14:28
 |Â
show 1 more comment
up vote
-1
down vote
I am surprised nobody has suggested using arrays. Here is my crude attempt (using the log3.txt idea from @Stephane above.
#!/bin/bash
nl1=$( wc -l < log1.txt )
nl2=$( wc -l < log2.txt )
nlnums=$( wc -l < nums.txt )
declare -a arr1[$nl1]
declare -a arr2[$nl2]
declare -a nums[$nlnums]
for (( i=0; i < $nl1; i++ ))
do
read arr1[$i]
done < log1.txt
for (( i=0; i < $nl2; i++ ))
do
read arr2[$i]
done < log2.txt
for (( i=0; i < $nlnums; i++ ))
do
read nums[$i]
done < nums.txt
j=0
for (( i=0; i < $nl1; i++ ))
do
echo "The $nums[$i] color $arr1[$i] is $arr2[$j]"
j=$(( (j+1) % $nl2 ))
done
The file nums.txt is as follows:
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
The code needs to be cleaned a bit but illustrates the point.
answered Sep 7 at 22:32
unxnut
3,3802918
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
add a comment |Â
up vote
-1
down vote
I am surprised nobody has suggested using arrays. Here is my crude attempt (using the log3.txt idea from @Stephane above.
#!/bin/bash
nl1=$( wc -l < log1.txt )
nl2=$( wc -l < log2.txt )
nlnums=$( wc -l < nums.txt )
declare -a arr1[$nl1]
declare -a arr2[$nl2]
declare -a nums[$nlnums]
for (( i=0; i < $nl1; i++ ))
do
read arr1[$i]
done < log1.txt
for (( i=0; i < $nl2; i++ ))
do
read arr2[$i]
done < log2.txt
for (( i=0; i < $nlnums; i++ ))
do
read nums[$i]
done < nums.txt
j=0
for (( i=0; i < $nl1; i++ ))
do
echo "The $nums[$i] color $arr1[$i] is $arr2[$j]"
j=$(( (j+1) % $nl2 ))
done
The file nums.txt is as follows:
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
The code needs to be cleaned a bit but illustrates the point.
answered Sep 7 at 22:32
unxnut
3,3802918
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
I am surprised nobody has suggested using arrays. Here is my crude attempt (using the log3.txt idea from @Stephane above.
#!/bin/bash
nl1=$( wc -l < log1.txt )
nl2=$( wc -l < log2.txt )
nlnums=$( wc -l < nums.txt )
declare -a arr1[$nl1]
declare -a arr2[$nl2]
declare -a nums[$nlnums]
for (( i=0; i < $nl1; i++ ))
do
read arr1[$i]
done < log1.txt
for (( i=0; i < $nl2; i++ ))
do
read arr2[$i]
done < log2.txt
for (( i=0; i < $nlnums; i++ ))
do
read nums[$i]
done < nums.txt
j=0
for (( i=0; i < $nl1; i++ ))
do
echo "The $nums[$i] color $arr1[$i] is $arr2[$j]"
j=$(( (j+1) % $nl2 ))
done
The file nums.txt is as follows:
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
The code needs to be cleaned a bit but illustrates the point.
answered Sep 7 at 22:32
unxnut
3,3802918
I am surprised nobody has suggested using arrays. Here is my crude attempt (using the log3.txt idea from @Stephane above.
#!/bin/bash
nl1=$( wc -l < log1.txt )
nl2=$( wc -l < log2.txt )
nlnums=$( wc -l < nums.txt )
declare -a arr1[$nl1]
declare -a arr2[$nl2]
declare -a nums[$nlnums]
for (( i=0; i < $nl1; i++ ))
do
read arr1[$i]
done < log1.txt
for (( i=0; i < $nl2; i++ ))
do
read arr2[$i]
done < log2.txt
for (( i=0; i < $nlnums; i++ ))
do
read nums[$i]
done < nums.txt
j=0
for (( i=0; i < $nl1; i++ ))
do
echo "The $nums[$i] color $arr1[$i] is $arr2[$j]"
j=$(( (j+1) % $nl2 ))
done
The file nums.txt is as follows:
first
second
third
fourth
fifth
sixth
seventh
eighth
ninth
tenth
The code needs to be cleaned a bit but illustrates the point.
answered Sep 7 at 22:32
unxnut
3,3802918
edited Sep 8 at 1:47
answered Sep 7 at 22:32
unxnut
3,3802918
answered Sep 7 at 22:32
unxnut
3,3802918
answered Sep 7 at 22:32
unxnut
3,3802918
3,3802918
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
add a comment |Â
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
Mr. Downvoter, can you please explain why you downvoted?
– unxnut
Sep 8 at 1:43
add a comment |Â
Kasper is a new contributor. Be nice, and check out our Code of Conduct.
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Kasper is a new contributor. Be nice, and check out our Code of Conduct.
Kasper is a new contributor. Be nice, and check out our Code of Conduct.
Kasper is a new contributor. Be nice, and check out our Code of Conduct.
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Post what you've tried so far, so we can see what tweak it might need ?
– steve
Sep 7 at 13:15
The name of the operation that you are performing on two tables in that script is a " join", for reference. However, what you actually want is not.
– JdeBP
Sep 7 at 13:42
2
What happened to seventh?
– Stéphane Chazelas
Sep 7 at 13:44
1
Would it be appropriate to ask why you want to do this in bash, rather than simply using a more capable scripting language?
– jamesqf
Sep 7 at 17:22