Mixing
List comprehension to extract multiple fields from list of tuples
Clash Royale CLAN TAG#URR8PPP
up vote
6
down vote
favorite
I have a list of tuples
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
I want to create a new list that only has the first two fields as in:
[('server1', 80), ('server2', 443)]
but I cannot see how to craft a list comprehension for more than one element.
hosts = [x[0] for x in servers] # this works to give me ['server1', server2']
hostswithports = [x[0], x[1] for x in servers] # this does not work
I prefer to learn the pythonic way vs using a loop - what am I doing wrong?
python tuples itertools
asked 9 hours ago
Bill
6116
add a comment |Â
up vote
6
down vote
favorite
I have a list of tuples
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
I want to create a new list that only has the first two fields as in:
[('server1', 80), ('server2', 443)]
but I cannot see how to craft a list comprehension for more than one element.
hosts = [x[0] for x in servers] # this works to give me ['server1', server2']
hostswithports = [x[0], x[1] for x in servers] # this does not work
I prefer to learn the pythonic way vs using a loop - what am I doing wrong?
python tuples itertools
asked 9 hours ago
Bill
6116
2
hostswithports = [ (x[0], x[1]) for x in servers]does it
– Patrick Artner
9 hours ago
3
loops are pythonic
– juanpa.arrivillaga
9 hours ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I have a list of tuples
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
I want to create a new list that only has the first two fields as in:
[('server1', 80), ('server2', 443)]
but I cannot see how to craft a list comprehension for more than one element.
hosts = [x[0] for x in servers] # this works to give me ['server1', server2']
hostswithports = [x[0], x[1] for x in servers] # this does not work
I prefer to learn the pythonic way vs using a loop - what am I doing wrong?
python tuples itertools
asked 9 hours ago
Bill
6116
I have a list of tuples
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
I want to create a new list that only has the first two fields as in:
[('server1', 80), ('server2', 443)]
but I cannot see how to craft a list comprehension for more than one element.
hosts = [x[0] for x in servers] # this works to give me ['server1', server2']
hostswithports = [x[0], x[1] for x in servers] # this does not work
I prefer to learn the pythonic way vs using a loop - what am I doing wrong?
python tuples itertools
python tuples itertools
asked 9 hours ago
Bill
6116
asked 9 hours ago
Bill
6116
asked 9 hours ago
Bill
6116
asked 9 hours ago
Bill
6116
asked 9 hours ago
Bill
6116
6116
2
hostswithports = [ (x[0], x[1]) for x in servers]does it
– Patrick Artner
9 hours ago
3
loops are pythonic
– juanpa.arrivillaga
9 hours ago
add a comment |Â
2
hostswithports = [ (x[0], x[1]) for x in servers]does it
– Patrick Artner
9 hours ago
3
loops are pythonic
– juanpa.arrivillaga
9 hours ago
2
2
hostswithports = [ (x[0], x[1]) for x in servers] does it– Patrick Artner
9 hours ago
hostswithports = [ (x[0], x[1]) for x in servers] does it– Patrick Artner
9 hours ago
3
3
loops are pythonic
– juanpa.arrivillaga
9 hours ago
loops are pythonic
– juanpa.arrivillaga
9 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
13
down vote
accepted
You can use extended iterable unpacking.
>>> servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
>>> [(server, port) for server, port, *_ in servers]
>>> [('server1', 80), ('server2', 443)]
Using _ as a throwaway placeholder-name is a common convention.
answered 9 hours ago
timgeb
40.6k105580
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
... and for older versions,[(server, port) for server, port, _, _ in servers]works just as well.
– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:a, b, *_will accept elements with at least two sub-elements.a, b, _, _only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)
– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
add a comment |Â
up vote
4
down vote
Using basic slicing, which has the benefit of not failing if any of your list elements don't have the expected number of sub-elements.
[el[:2] for el in servers]
[('server1', 80), ('server2', 443)]
answered 9 hours ago
user3483203
27.7k72350
Suggestion:server_port = slice(2)->[el[server_port] for ...]would make for nice self-documentation.
– timgeb
9 hours ago
add a comment |Â
up vote
3
down vote
You could use itemgetter:
from operator import itemgetter
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
result = [e for e in map(itemgetter(0, 1), servers)]
print(result)
Output
[('server1', 80), ('server2', 443)]
A more readable alternative is the following:
from operator import itemgetter
get_server_and_port = itemgetter(0, 1)
servers = [('server1', 80, 1, 2), ('server2', 443, 3, 4)]
result = [get_server_and_port(e) for e in servers]
print(result) # [('server1', 80), ('server2', 443)]
answered 9 hours ago
Daniel Mesejo
5,338620
add a comment |Â
up vote
2
down vote
What you were doing was almost right. You were attempting to translate each tuple in your list into a new tuple. But you forgot to actually declare the tuple. That's what the parentheses are doing:
hosts = [(x[0], x[1]) for x in servers]
answered 9 hours ago
Woody1193
1,497725
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
down vote
accepted
You can use extended iterable unpacking.
>>> servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
>>> [(server, port) for server, port, *_ in servers]
>>> [('server1', 80), ('server2', 443)]
Using _ as a throwaway placeholder-name is a common convention.
answered 9 hours ago
timgeb
40.6k105580
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
... and for older versions,[(server, port) for server, port, _, _ in servers]works just as well.
– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:a, b, *_will accept elements with at least two sub-elements.a, b, _, _only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)
– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
add a comment |Â
up vote
13
down vote
accepted
You can use extended iterable unpacking.
>>> servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
>>> [(server, port) for server, port, *_ in servers]
>>> [('server1', 80), ('server2', 443)]
Using _ as a throwaway placeholder-name is a common convention.
answered 9 hours ago
timgeb
40.6k105580
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
... and for older versions,[(server, port) for server, port, _, _ in servers]works just as well.
– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:a, b, *_will accept elements with at least two sub-elements.a, b, _, _only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)
– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
add a comment |Â
up vote
13
down vote
accepted
up vote
13
down vote
accepted
You can use extended iterable unpacking.
>>> servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
>>> [(server, port) for server, port, *_ in servers]
>>> [('server1', 80), ('server2', 443)]
Using _ as a throwaway placeholder-name is a common convention.
answered 9 hours ago
timgeb
40.6k105580
You can use extended iterable unpacking.
>>> servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
>>> [(server, port) for server, port, *_ in servers]
>>> [('server1', 80), ('server2', 443)]
Using _ as a throwaway placeholder-name is a common convention.
answered 9 hours ago
timgeb
40.6k105580
edited 9 hours ago
answered 9 hours ago
timgeb
40.6k105580
answered 9 hours ago
timgeb
40.6k105580
answered 9 hours ago
timgeb
40.6k105580
40.6k105580
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
... and for older versions,[(server, port) for server, port, _, _ in servers]works just as well.
– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:a, b, *_will accept elements with at least two sub-elements.a, b, _, _only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)
– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
add a comment |Â
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
... and for older versions,[(server, port) for server, port, _, _ in servers]works just as well.
– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:a, b, *_will accept elements with at least two sub-elements.a, b, _, _only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)
– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
2
2
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
This is the only one that, by only looking at the line in question, I have any idea what is happening. Readability matters: +1
– TemporalWolf
9 hours ago
1
1
... and for older versions,
[(server, port) for server, port, _, _ in servers] works just as well.– tobias_k
9 hours ago
... and for older versions,
[(server, port) for server, port, _, _ in servers] works just as well.– tobias_k
9 hours ago
@tobias_k it is worth noting it handles things differently:
a, b, *_ will accept elements with at least two sub-elements. a, b, _, _ only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)– TemporalWolf
8 hours ago
@tobias_k it is worth noting it handles things differently:
a, b, *_ will accept elements with at least two sub-elements. a, b, _, _ only accepts elements with 4 sub-elements. So they are different in how they handle data that doesn't conform to the example. But with conforming data, they are the same... and in Py2.7 you don't get the choice ;)– TemporalWolf
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
I see I was just missing enclosing parentheses but I really like the readability of this approach
– Bill
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
@TemporalWolf Of course, I thought that was clear, but you are right: OP actually never says anything about the shape of the lists, just that the first two elements are needed, so it might indeed be irregular.
– tobias_k
8 hours ago
add a comment |Â
up vote
4
down vote
Using basic slicing, which has the benefit of not failing if any of your list elements don't have the expected number of sub-elements.
[el[:2] for el in servers]
[('server1', 80), ('server2', 443)]
answered 9 hours ago
user3483203
27.7k72350
Suggestion:server_port = slice(2)->[el[server_port] for ...]would make for nice self-documentation.
– timgeb
9 hours ago
add a comment |Â
up vote
4
down vote
Using basic slicing, which has the benefit of not failing if any of your list elements don't have the expected number of sub-elements.
[el[:2] for el in servers]
[('server1', 80), ('server2', 443)]
answered 9 hours ago
user3483203
27.7k72350
Suggestion:server_port = slice(2)->[el[server_port] for ...]would make for nice self-documentation.
– timgeb
9 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Using basic slicing, which has the benefit of not failing if any of your list elements don't have the expected number of sub-elements.
[el[:2] for el in servers]
[('server1', 80), ('server2', 443)]
answered 9 hours ago
user3483203
27.7k72350
Using basic slicing, which has the benefit of not failing if any of your list elements don't have the expected number of sub-elements.
[el[:2] for el in servers]
[('server1', 80), ('server2', 443)]
answered 9 hours ago
user3483203
27.7k72350
edited 9 hours ago
answered 9 hours ago
user3483203
27.7k72350
answered 9 hours ago
user3483203
27.7k72350
answered 9 hours ago
user3483203
27.7k72350
27.7k72350
Suggestion:server_port = slice(2)->[el[server_port] for ...]would make for nice self-documentation.
– timgeb
9 hours ago
add a comment |Â
Suggestion:server_port = slice(2)->[el[server_port] for ...]would make for nice self-documentation.
– timgeb
9 hours ago
Suggestion:
server_port = slice(2) -> [el[server_port] for ...] would make for nice self-documentation.– timgeb
9 hours ago
Suggestion:
server_port = slice(2) -> [el[server_port] for ...] would make for nice self-documentation.– timgeb
9 hours ago
add a comment |Â
up vote
3
down vote
You could use itemgetter:
from operator import itemgetter
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
result = [e for e in map(itemgetter(0, 1), servers)]
print(result)
Output
[('server1', 80), ('server2', 443)]
A more readable alternative is the following:
from operator import itemgetter
get_server_and_port = itemgetter(0, 1)
servers = [('server1', 80, 1, 2), ('server2', 443, 3, 4)]
result = [get_server_and_port(e) for e in servers]
print(result) # [('server1', 80), ('server2', 443)]
answered 9 hours ago
Daniel Mesejo
5,338620
add a comment |Â
up vote
3
down vote
You could use itemgetter:
from operator import itemgetter
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
result = [e for e in map(itemgetter(0, 1), servers)]
print(result)
Output
[('server1', 80), ('server2', 443)]
A more readable alternative is the following:
from operator import itemgetter
get_server_and_port = itemgetter(0, 1)
servers = [('server1', 80, 1, 2), ('server2', 443, 3, 4)]
result = [get_server_and_port(e) for e in servers]
print(result) # [('server1', 80), ('server2', 443)]
answered 9 hours ago
Daniel Mesejo
5,338620
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You could use itemgetter:
from operator import itemgetter
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
result = [e for e in map(itemgetter(0, 1), servers)]
print(result)
Output
[('server1', 80), ('server2', 443)]
A more readable alternative is the following:
from operator import itemgetter
get_server_and_port = itemgetter(0, 1)
servers = [('server1', 80, 1, 2), ('server2', 443, 3, 4)]
result = [get_server_and_port(e) for e in servers]
print(result) # [('server1', 80), ('server2', 443)]
answered 9 hours ago
Daniel Mesejo
5,338620
You could use itemgetter:
from operator import itemgetter
servers = [('server1', 80 , 1, 2), ('server2', 443, 3, 4)]
result = [e for e in map(itemgetter(0, 1), servers)]
print(result)
Output
[('server1', 80), ('server2', 443)]
A more readable alternative is the following:
from operator import itemgetter
get_server_and_port = itemgetter(0, 1)
servers = [('server1', 80, 1, 2), ('server2', 443, 3, 4)]
result = [get_server_and_port(e) for e in servers]
print(result) # [('server1', 80), ('server2', 443)]
answered 9 hours ago
Daniel Mesejo
5,338620
edited 9 hours ago
answered 9 hours ago
Daniel Mesejo
5,338620
answered 9 hours ago
Daniel Mesejo
5,338620
answered 9 hours ago
Daniel Mesejo
5,338620
5,338620
add a comment |Â
add a comment |Â
up vote
2
down vote
What you were doing was almost right. You were attempting to translate each tuple in your list into a new tuple. But you forgot to actually declare the tuple. That's what the parentheses are doing:
hosts = [(x[0], x[1]) for x in servers]
answered 9 hours ago
Woody1193
1,497725
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
add a comment |Â
up vote
2
down vote
What you were doing was almost right. You were attempting to translate each tuple in your list into a new tuple. But you forgot to actually declare the tuple. That's what the parentheses are doing:
hosts = [(x[0], x[1]) for x in servers]
answered 9 hours ago
Woody1193
1,497725
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
What you were doing was almost right. You were attempting to translate each tuple in your list into a new tuple. But you forgot to actually declare the tuple. That's what the parentheses are doing:
hosts = [(x[0], x[1]) for x in servers]
answered 9 hours ago
Woody1193
1,497725
What you were doing was almost right. You were attempting to translate each tuple in your list into a new tuple. But you forgot to actually declare the tuple. That's what the parentheses are doing:
hosts = [(x[0], x[1]) for x in servers]
answered 9 hours ago
Woody1193
1,497725
answered 9 hours ago
Woody1193
1,497725
answered 9 hours ago
Woody1193
1,497725
answered 9 hours ago
Woody1193
1,497725
1,497725
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
add a comment |Â
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
Thanks - I could have sworn I tried that too but obviously I did not!
– Bill
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
@Bill If I had a nickel for every time I did that . . . I'd be able to buy a steak dinner at least
– Woody1193
8 hours ago
add a comment |Â
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2
hostswithports = [ (x[0], x[1]) for x in servers]does it– Patrick Artner
9 hours ago
3
loops are pythonic
– juanpa.arrivillaga
9 hours ago