Need a defaultdict with values as two lists
up vote
0
down vote
favorite
In Python, I want something like
dict = defaultdict((list,list))
Essentially, with every key I want two lists!
With the above snippet I get the error first argument must be callable. How can I achieve this ?
python dictionary defaultdict
asked Nov 11 at 18:12
midi
7110
add a comment |
up vote
0
down vote
favorite
In Python, I want something like
dict = defaultdict((list,list))
Essentially, with every key I want two lists!
With the above snippet I get the error first argument must be callable. How can I achieve this ?
python dictionary defaultdict
asked Nov 11 at 18:12
midi
7110
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In Python, I want something like
dict = defaultdict((list,list))
Essentially, with every key I want two lists!
With the above snippet I get the error first argument must be callable. How can I achieve this ?
python dictionary defaultdict
asked Nov 11 at 18:12
midi
7110
In Python, I want something like
dict = defaultdict((list,list))
Essentially, with every key I want two lists!
With the above snippet I get the error first argument must be callable. How can I achieve this ?
python dictionary defaultdict
python dictionary defaultdict
asked Nov 11 at 18:12
midi
7110
asked Nov 11 at 18:12
midi
7110
asked Nov 11 at 18:12
midi
7110
asked Nov 11 at 18:12
midi
7110
asked Nov 11 at 18:12
midi
7110
7110
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22
add a comment |
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Give as parameter to defaultdict a function that creates your empty lists:
from collections import defaultdict
def pair_of_lists():
return [, ]
d = defaultdict(pair_of_lists)
d[1][0].append(3)
d[1][1].append(42)
print(d)
# defaultdict(<function pair_of_lists at 0x7f584a40b0d0>, {1: [[3], [42]]})
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
add a comment |
up vote
0
down vote
It is not some kind of type inference, you just provide a function that generates default value. While int, list, dict without argument generate 0, , {}, which is often exploited in defaultdict declaration. Python has no build in constructor function for pair of list etc. So it is as simple as
di = defaultdict(lambda : ([1,2,3], ['a', 'cb']))
While Python allows tuples (pairs) of lists, you might have some issues e.g. with hashing of list tuples. To be safe, you can set deafult to list of two list. Note you have set it to specific lists, empty or not, and not list type/constructor. It is literally about setting default value and not a type declaration - python lists are untyped.
from collections import defaultdict
l1 =
l2 = ["another", "default", "list"]
di = defaultdict(lambda : [ l1, l2] )
print (di[3])
answered Nov 11 at 18:15
Serge
1,366522
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :dict = defaultdict(lambda : [, ])but this does notdict = defaultdict(lambda : [list, list])Cannot figure out why..
– midi
Nov 11 at 18:27
@midi where you writelist, it doesn’t get called so you should write eitherorlist()
– N Chauhan
Nov 11 at 18:43
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53251716%2fneed-a-defaultdict-with-values-as-two-lists%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Give as parameter to defaultdict a function that creates your empty lists:
from collections import defaultdict
def pair_of_lists():
return [, ]
d = defaultdict(pair_of_lists)
d[1][0].append(3)
d[1][1].append(42)
print(d)
# defaultdict(<function pair_of_lists at 0x7f584a40b0d0>, {1: [[3], [42]]})
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
add a comment |
up vote
1
down vote
accepted
Give as parameter to defaultdict a function that creates your empty lists:
from collections import defaultdict
def pair_of_lists():
return [, ]
d = defaultdict(pair_of_lists)
d[1][0].append(3)
d[1][1].append(42)
print(d)
# defaultdict(<function pair_of_lists at 0x7f584a40b0d0>, {1: [[3], [42]]})
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Give as parameter to defaultdict a function that creates your empty lists:
from collections import defaultdict
def pair_of_lists():
return [, ]
d = defaultdict(pair_of_lists)
d[1][0].append(3)
d[1][1].append(42)
print(d)
# defaultdict(<function pair_of_lists at 0x7f584a40b0d0>, {1: [[3], [42]]})
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
Give as parameter to defaultdict a function that creates your empty lists:
from collections import defaultdict
def pair_of_lists():
return [, ]
d = defaultdict(pair_of_lists)
d[1][0].append(3)
d[1][1].append(42)
print(d)
# defaultdict(<function pair_of_lists at 0x7f584a40b0d0>, {1: [[3], [42]]})
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
answered Nov 11 at 18:18
Thierry Lathuille
7,12182630
7,12182630
add a comment |
add a comment |
up vote
0
down vote
It is not some kind of type inference, you just provide a function that generates default value. While int, list, dict without argument generate 0, , {}, which is often exploited in defaultdict declaration. Python has no build in constructor function for pair of list etc. So it is as simple as
di = defaultdict(lambda : ([1,2,3], ['a', 'cb']))
While Python allows tuples (pairs) of lists, you might have some issues e.g. with hashing of list tuples. To be safe, you can set deafult to list of two list. Note you have set it to specific lists, empty or not, and not list type/constructor. It is literally about setting default value and not a type declaration - python lists are untyped.
from collections import defaultdict
l1 =
l2 = ["another", "default", "list"]
di = defaultdict(lambda : [ l1, l2] )
print (di[3])
answered Nov 11 at 18:15
Serge
1,366522
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :dict = defaultdict(lambda : [, ])but this does notdict = defaultdict(lambda : [list, list])Cannot figure out why..
– midi
Nov 11 at 18:27
@midi where you writelist, it doesn’t get called so you should write eitherorlist()
– N Chauhan
Nov 11 at 18:43
add a comment |
up vote
0
down vote
It is not some kind of type inference, you just provide a function that generates default value. While int, list, dict without argument generate 0, , {}, which is often exploited in defaultdict declaration. Python has no build in constructor function for pair of list etc. So it is as simple as
di = defaultdict(lambda : ([1,2,3], ['a', 'cb']))
While Python allows tuples (pairs) of lists, you might have some issues e.g. with hashing of list tuples. To be safe, you can set deafult to list of two list. Note you have set it to specific lists, empty or not, and not list type/constructor. It is literally about setting default value and not a type declaration - python lists are untyped.
from collections import defaultdict
l1 =
l2 = ["another", "default", "list"]
di = defaultdict(lambda : [ l1, l2] )
print (di[3])
answered Nov 11 at 18:15
Serge
1,366522
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :dict = defaultdict(lambda : [, ])but this does notdict = defaultdict(lambda : [list, list])Cannot figure out why..
– midi
Nov 11 at 18:27
@midi where you writelist, it doesn’t get called so you should write eitherorlist()
– N Chauhan
Nov 11 at 18:43
add a comment |
up vote
0
down vote
up vote
0
down vote
It is not some kind of type inference, you just provide a function that generates default value. While int, list, dict without argument generate 0, , {}, which is often exploited in defaultdict declaration. Python has no build in constructor function for pair of list etc. So it is as simple as
di = defaultdict(lambda : ([1,2,3], ['a', 'cb']))
While Python allows tuples (pairs) of lists, you might have some issues e.g. with hashing of list tuples. To be safe, you can set deafult to list of two list. Note you have set it to specific lists, empty or not, and not list type/constructor. It is literally about setting default value and not a type declaration - python lists are untyped.
from collections import defaultdict
l1 =
l2 = ["another", "default", "list"]
di = defaultdict(lambda : [ l1, l2] )
print (di[3])
answered Nov 11 at 18:15
Serge
1,366522
It is not some kind of type inference, you just provide a function that generates default value. While int, list, dict without argument generate 0, , {}, which is often exploited in defaultdict declaration. Python has no build in constructor function for pair of list etc. So it is as simple as
di = defaultdict(lambda : ([1,2,3], ['a', 'cb']))
While Python allows tuples (pairs) of lists, you might have some issues e.g. with hashing of list tuples. To be safe, you can set deafult to list of two list. Note you have set it to specific lists, empty or not, and not list type/constructor. It is literally about setting default value and not a type declaration - python lists are untyped.
from collections import defaultdict
l1 =
l2 = ["another", "default", "list"]
di = defaultdict(lambda : [ l1, l2] )
print (di[3])
answered Nov 11 at 18:15
Serge
1,366522
edited Nov 12 at 21:20
answered Nov 11 at 18:15
Serge
1,366522
answered Nov 11 at 18:15
Serge
1,366522
answered Nov 11 at 18:15
Serge
1,366522
1,366522
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :dict = defaultdict(lambda : [, ])but this does notdict = defaultdict(lambda : [list, list])Cannot figure out why..
– midi
Nov 11 at 18:27
@midi where you writelist, it doesn’t get called so you should write eitherorlist()
– N Chauhan
Nov 11 at 18:43
add a comment |
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :dict = defaultdict(lambda : [, ])but this does notdict = defaultdict(lambda : [list, list])Cannot figure out why..
– midi
Nov 11 at 18:27
@midi where you writelist, it doesn’t get called so you should write eitherorlist()
– N Chauhan
Nov 11 at 18:43
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
Strangely, dict['This'][0].append(1) does not work! what am I missing ?
– midi
Nov 11 at 18:23
This works :
dict = defaultdict(lambda : [, ]) but this does not dict = defaultdict(lambda : [list, list]) Cannot figure out why..– midi
Nov 11 at 18:27
This works :
dict = defaultdict(lambda : [, ]) but this does not dict = defaultdict(lambda : [list, list]) Cannot figure out why..– midi
Nov 11 at 18:27
@midi where you write
list, it doesn’t get called so you should write either list()– N Chauhan
Nov 11 at 18:43
@midi where you write
list, it doesn’t get called so you should write either list()– N Chauhan
Nov 11 at 18:43
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53251716%2fneed-a-defaultdict-with-values-as-two-lists%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
@Austin, this would make the code a bit untidy, as after every new key insert, I would have to first append two lists.. extra if checks here..
– midi
Nov 11 at 18:22