Mixing
Better method to iterate over 3 lists
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I am creating a program which iterates over the width and height of an image as well as makes use of a set of keys.
Here is an example:
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
The width and height are a range of integers up to the size of the width and height.
The keys are any set of numbers (actually ASCII values) but are not ordered numbers.
I would like the output to be like this:
0 0 18
0 1 20
0 2 11
0 3 18
0 4 20
1 0 11
1 1 18
. . ..
9 0 20
9 1 11
9 2 18
9 3 20
9 4 11
As you can see, the width and height can be produced with nested for loops, whereas the keys cycle between each other.
Here is my solution:
w = [0,1,2,3,4,6,7,8,9]
h = [0,1,2,3,4]
k = [18,20,11]
kIndex = 0
for i in w:
for j in h:
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
Actually it works as intended, however, I would like a more efficient way to do the above instead of using an incremental variable for the index position of the keys list.
I don't mind the nested for loop, if that has to be used, but the index keys variable annoys me as it seems like the code won't work without it, but at the same time isn't really pythonic.
python
asked 4 hours ago
Suraj Kothari
300112
add a comment |Â
up vote
6
down vote
favorite
I am creating a program which iterates over the width and height of an image as well as makes use of a set of keys.
Here is an example:
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
The width and height are a range of integers up to the size of the width and height.
The keys are any set of numbers (actually ASCII values) but are not ordered numbers.
I would like the output to be like this:
0 0 18
0 1 20
0 2 11
0 3 18
0 4 20
1 0 11
1 1 18
. . ..
9 0 20
9 1 11
9 2 18
9 3 20
9 4 11
As you can see, the width and height can be produced with nested for loops, whereas the keys cycle between each other.
Here is my solution:
w = [0,1,2,3,4,6,7,8,9]
h = [0,1,2,3,4]
k = [18,20,11]
kIndex = 0
for i in w:
for j in h:
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
Actually it works as intended, however, I would like a more efficient way to do the above instead of using an incremental variable for the index position of the keys list.
I don't mind the nested for loop, if that has to be used, but the index keys variable annoys me as it seems like the code won't work without it, but at the same time isn't really pythonic.
python
asked 4 hours ago
Suraj Kothari
300112
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I am creating a program which iterates over the width and height of an image as well as makes use of a set of keys.
Here is an example:
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
The width and height are a range of integers up to the size of the width and height.
The keys are any set of numbers (actually ASCII values) but are not ordered numbers.
I would like the output to be like this:
0 0 18
0 1 20
0 2 11
0 3 18
0 4 20
1 0 11
1 1 18
. . ..
9 0 20
9 1 11
9 2 18
9 3 20
9 4 11
As you can see, the width and height can be produced with nested for loops, whereas the keys cycle between each other.
Here is my solution:
w = [0,1,2,3,4,6,7,8,9]
h = [0,1,2,3,4]
k = [18,20,11]
kIndex = 0
for i in w:
for j in h:
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
Actually it works as intended, however, I would like a more efficient way to do the above instead of using an incremental variable for the index position of the keys list.
I don't mind the nested for loop, if that has to be used, but the index keys variable annoys me as it seems like the code won't work without it, but at the same time isn't really pythonic.
python
asked 4 hours ago
Suraj Kothari
300112
I am creating a program which iterates over the width and height of an image as well as makes use of a set of keys.
Here is an example:
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
The width and height are a range of integers up to the size of the width and height.
The keys are any set of numbers (actually ASCII values) but are not ordered numbers.
I would like the output to be like this:
0 0 18
0 1 20
0 2 11
0 3 18
0 4 20
1 0 11
1 1 18
. . ..
9 0 20
9 1 11
9 2 18
9 3 20
9 4 11
As you can see, the width and height can be produced with nested for loops, whereas the keys cycle between each other.
Here is my solution:
w = [0,1,2,3,4,6,7,8,9]
h = [0,1,2,3,4]
k = [18,20,11]
kIndex = 0
for i in w:
for j in h:
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
Actually it works as intended, however, I would like a more efficient way to do the above instead of using an incremental variable for the index position of the keys list.
I don't mind the nested for loop, if that has to be used, but the index keys variable annoys me as it seems like the code won't work without it, but at the same time isn't really pythonic.
python
python
asked 4 hours ago
Suraj Kothari
300112
asked 4 hours ago
Suraj Kothari
300112
asked 4 hours ago
Suraj Kothari
300112
asked 4 hours ago
Suraj Kothari
300112
asked 4 hours ago
Suraj Kothari
300112
300112
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
You can use itertools.product to get the product of you width and height. You then want to cycle over the keys, thus use itertools.cycle. You then zip those together and get the desired result.
Finally, you can make this a generator using yield for memory efficieny.
from itertools import product, cycle
def get_grid(width, height, keys):
for pos, key in zip(product(width, height), cycle(keys)):
yield (*pos, key)
Or if you do not want a generator.
out = [(*pos, key) for pos, key in zip(product(width, height), cycle(keys))]
Example
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
for triple in get_grid(width, height, keys):
print(triple)
Output
(0, 0, 18)
(0, 1, 20)
(0, 2, 11)
(0, 3, 18)
(0, 4, 20)
(1, 0, 11)
(1, 1, 18)
...
As a sidenote, notice that you could replace the lists defining width and height by ranges.
width = range(10)
height = range(5)
answered 4 hours ago
Olivier Melançon
12.3k11636
1
Pretty much what I was about to post:[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]
– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to alistby doinglist(get_grid(width, height, keys)).
– Olivier Melançon
3 hours ago
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
 |Â
show 2 more comments
up vote
1
down vote
You can use cycle from itertools.
from itertools import cycle, product
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
c = cycle(keys)
for w,h in product(width,height):
print(w,h,next(c))
answered 3 hours ago
Christian Sloper
6817
add a comment |Â
up vote
0
down vote
You can use iterate over matrix. Its ndenumerate function from numpy packet. If width and height are range of int, you can skip creating list. Just define the size of this lists.
width = 10
height = 5
k = [18,20,11]
kIndex = 0
for (i,j), value in np.ndenumerate(np.ones((height,width))):
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
answered 3 hours ago
Cezary.Sz
587414
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
You can use itertools.product to get the product of you width and height. You then want to cycle over the keys, thus use itertools.cycle. You then zip those together and get the desired result.
Finally, you can make this a generator using yield for memory efficieny.
from itertools import product, cycle
def get_grid(width, height, keys):
for pos, key in zip(product(width, height), cycle(keys)):
yield (*pos, key)
Or if you do not want a generator.
out = [(*pos, key) for pos, key in zip(product(width, height), cycle(keys))]
Example
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
for triple in get_grid(width, height, keys):
print(triple)
Output
(0, 0, 18)
(0, 1, 20)
(0, 2, 11)
(0, 3, 18)
(0, 4, 20)
(1, 0, 11)
(1, 1, 18)
...
As a sidenote, notice that you could replace the lists defining width and height by ranges.
width = range(10)
height = range(5)
answered 4 hours ago
Olivier Melançon
12.3k11636
1
Pretty much what I was about to post:[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]
– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to alistby doinglist(get_grid(width, height, keys)).
– Olivier Melançon
3 hours ago
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
 |Â
show 2 more comments
up vote
9
down vote
accepted
You can use itertools.product to get the product of you width and height. You then want to cycle over the keys, thus use itertools.cycle. You then zip those together and get the desired result.
Finally, you can make this a generator using yield for memory efficieny.
from itertools import product, cycle
def get_grid(width, height, keys):
for pos, key in zip(product(width, height), cycle(keys)):
yield (*pos, key)
Or if you do not want a generator.
out = [(*pos, key) for pos, key in zip(product(width, height), cycle(keys))]
Example
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
for triple in get_grid(width, height, keys):
print(triple)
Output
(0, 0, 18)
(0, 1, 20)
(0, 2, 11)
(0, 3, 18)
(0, 4, 20)
(1, 0, 11)
(1, 1, 18)
...
As a sidenote, notice that you could replace the lists defining width and height by ranges.
width = range(10)
height = range(5)
answered 4 hours ago
Olivier Melançon
12.3k11636
1
Pretty much what I was about to post:[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]
– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to alistby doinglist(get_grid(width, height, keys)).
– Olivier Melançon
3 hours ago
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
 |Â
show 2 more comments
up vote
9
down vote
accepted
up vote
9
down vote
accepted
You can use itertools.product to get the product of you width and height. You then want to cycle over the keys, thus use itertools.cycle. You then zip those together and get the desired result.
Finally, you can make this a generator using yield for memory efficieny.
from itertools import product, cycle
def get_grid(width, height, keys):
for pos, key in zip(product(width, height), cycle(keys)):
yield (*pos, key)
Or if you do not want a generator.
out = [(*pos, key) for pos, key in zip(product(width, height), cycle(keys))]
Example
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
for triple in get_grid(width, height, keys):
print(triple)
Output
(0, 0, 18)
(0, 1, 20)
(0, 2, 11)
(0, 3, 18)
(0, 4, 20)
(1, 0, 11)
(1, 1, 18)
...
As a sidenote, notice that you could replace the lists defining width and height by ranges.
width = range(10)
height = range(5)
answered 4 hours ago
Olivier Melançon
12.3k11636
You can use itertools.product to get the product of you width and height. You then want to cycle over the keys, thus use itertools.cycle. You then zip those together and get the desired result.
Finally, you can make this a generator using yield for memory efficieny.
from itertools import product, cycle
def get_grid(width, height, keys):
for pos, key in zip(product(width, height), cycle(keys)):
yield (*pos, key)
Or if you do not want a generator.
out = [(*pos, key) for pos, key in zip(product(width, height), cycle(keys))]
Example
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
for triple in get_grid(width, height, keys):
print(triple)
Output
(0, 0, 18)
(0, 1, 20)
(0, 2, 11)
(0, 3, 18)
(0, 4, 20)
(1, 0, 11)
(1, 1, 18)
...
As a sidenote, notice that you could replace the lists defining width and height by ranges.
width = range(10)
height = range(5)
answered 4 hours ago
Olivier Melançon
12.3k11636
edited 3 hours ago
answered 4 hours ago
Olivier Melançon
12.3k11636
answered 4 hours ago
Olivier Melançon
12.3k11636
answered 4 hours ago
Olivier Melançon
12.3k11636
12.3k11636
1
Pretty much what I was about to post:[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]
– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to alistby doinglist(get_grid(width, height, keys)).
– Olivier Melançon
3 hours ago
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
 |Â
show 2 more comments
1
Pretty much what I was about to post:[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]
– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to alistby doinglist(get_grid(width, height, keys)).
– Olivier Melançon
3 hours ago
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
1
1
Pretty much what I was about to post:
[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]– tobias_k
4 hours ago
Pretty much what I was about to post:
[(w,h,k) for (w,h),k in zip(product(width, height), cycle(keys))]– tobias_k
4 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
Is there a way without a generator. I'm not too confident with them. The itertools library seems ok, but the generator seems off my level.
– Suraj Kothari
3 hours ago
@SurajKothari Yes, you can simply cast the generator to a
list by doing list(get_grid(width, height, keys)).– Olivier Melançon
3 hours ago
@SurajKothari Yes, you can simply cast the generator to a
list by doing list(get_grid(width, height, keys)).– Olivier Melançon
3 hours ago
1
1
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
I have used width = range(Image.size[0]), but to understand my problem I used my example with an arbitrary list
– Suraj Kothari
3 hours ago
1
1
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
Thanks for the list comprehension. I seem to understand it slightly better since I use them all the time. However, it has dawned on me that the generator version is more "beautiful" or elegant and I might consider them anyways. Thanks a lot.
– Suraj Kothari
3 hours ago
 |Â
show 2 more comments
up vote
1
down vote
You can use cycle from itertools.
from itertools import cycle, product
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
c = cycle(keys)
for w,h in product(width,height):
print(w,h,next(c))
answered 3 hours ago
Christian Sloper
6817
add a comment |Â
up vote
1
down vote
You can use cycle from itertools.
from itertools import cycle, product
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
c = cycle(keys)
for w,h in product(width,height):
print(w,h,next(c))
answered 3 hours ago
Christian Sloper
6817
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You can use cycle from itertools.
from itertools import cycle, product
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
c = cycle(keys)
for w,h in product(width,height):
print(w,h,next(c))
answered 3 hours ago
Christian Sloper
6817
You can use cycle from itertools.
from itertools import cycle, product
width = [0,1,2,3,4,6,7,8,9]
height = [0,1,2,3,4]
keys = [18,20,11]
c = cycle(keys)
for w,h in product(width,height):
print(w,h,next(c))
answered 3 hours ago
Christian Sloper
6817
edited 3 hours ago
answered 3 hours ago
Christian Sloper
6817
answered 3 hours ago
Christian Sloper
6817
answered 3 hours ago
Christian Sloper
6817
6817
add a comment |Â
add a comment |Â
up vote
0
down vote
You can use iterate over matrix. Its ndenumerate function from numpy packet. If width and height are range of int, you can skip creating list. Just define the size of this lists.
width = 10
height = 5
k = [18,20,11]
kIndex = 0
for (i,j), value in np.ndenumerate(np.ones((height,width))):
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
answered 3 hours ago
Cezary.Sz
587414
add a comment |Â
up vote
0
down vote
You can use iterate over matrix. Its ndenumerate function from numpy packet. If width and height are range of int, you can skip creating list. Just define the size of this lists.
width = 10
height = 5
k = [18,20,11]
kIndex = 0
for (i,j), value in np.ndenumerate(np.ones((height,width))):
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
answered 3 hours ago
Cezary.Sz
587414
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You can use iterate over matrix. Its ndenumerate function from numpy packet. If width and height are range of int, you can skip creating list. Just define the size of this lists.
width = 10
height = 5
k = [18,20,11]
kIndex = 0
for (i,j), value in np.ndenumerate(np.ones((height,width))):
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
answered 3 hours ago
Cezary.Sz
587414
You can use iterate over matrix. Its ndenumerate function from numpy packet. If width and height are range of int, you can skip creating list. Just define the size of this lists.
width = 10
height = 5
k = [18,20,11]
kIndex = 0
for (i,j), value in np.ndenumerate(np.ones((height,width))):
print(i,j,k[kIndex])
# Cycle through the keys index.
# The modulo is used to return to the beginning of the keys list
kIndex = (kIndex + 1) % len(k)
answered 3 hours ago
Cezary.Sz
587414
answered 3 hours ago
Cezary.Sz
587414
answered 3 hours ago
Cezary.Sz
587414
answered 3 hours ago
Cezary.Sz
587414
587414
add a comment |Â
add a comment |Â
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