How to change the first value in the tuple of a list?
1
This is my matrix:
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5,
0.023)]]
And I want to let it to be a
b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10,
0.023)]]
To add n for the first value in the tuple, and I tried:
for n in b:
for ee,ww in n:
ee == ee + 2903
It doesn't work.
How should I keep the change to the original matrix b?
python numpy replace tuples
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
add a comment |
1
This is my matrix:
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5,
0.023)]]
And I want to let it to be a
b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10,
0.023)]]
To add n for the first value in the tuple, and I tried:
for n in b:
for ee,ww in n:
ee == ee + 2903
It doesn't work.
How should I keep the change to the original matrix b?
python numpy replace tuples
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
2
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
1
I don't understand where the2903comes from, andbis three levels deep while your for loop is only two. Also you're using the comparison==rather than the assignment=.
– Acccumulation
Nov 19 '18 at 16:34
If I gave you a functionupdate, would it have to updatebin place, so thatprint(b); update(b); print(b)would output two different values, or can it return a copynew_b = update(b). In the second case, do you need to retain the original object? That is, afternew_b = update(b), would you requireid(new_b) == id(b)? Would you require thatid(new_b[0]) == id(b[0])as well? (Keeping the existing rows, and only replacing the tuples within each row?)
– chepner
Nov 19 '18 at 16:38
ee== ..is not an assignment. Andee=..changes the variable but does not changen.
– hpaulj
Nov 19 '18 at 16:47
add a comment |
1
1
1
1
This is my matrix:
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5,
0.023)]]
And I want to let it to be a
b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10,
0.023)]]
To add n for the first value in the tuple, and I tried:
for n in b:
for ee,ww in n:
ee == ee + 2903
It doesn't work.
How should I keep the change to the original matrix b?
python numpy replace tuples
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
This is my matrix:
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5,
0.023)]]
And I want to let it to be a
b = [[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10,
0.023)]]
To add n for the first value in the tuple, and I tried:
for n in b:
for ee,ww in n:
ee == ee + 2903
It doesn't work.
How should I keep the change to the original matrix b?
python numpy replace tuples
python numpy replace tuples
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
edited Nov 19 '18 at 16:29
jdehesa
23.8k43554
23.8k43554
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
asked Nov 19 '18 at 16:27
賴韋安賴韋安
142
142
2
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
1
I don't understand where the2903comes from, andbis three levels deep while your for loop is only two. Also you're using the comparison==rather than the assignment=.
– Acccumulation
Nov 19 '18 at 16:34
If I gave you a functionupdate, would it have to updatebin place, so thatprint(b); update(b); print(b)would output two different values, or can it return a copynew_b = update(b). In the second case, do you need to retain the original object? That is, afternew_b = update(b), would you requireid(new_b) == id(b)? Would you require thatid(new_b[0]) == id(b[0])as well? (Keeping the existing rows, and only replacing the tuples within each row?)
– chepner
Nov 19 '18 at 16:38
ee== ..is not an assignment. Andee=..changes the variable but does not changen.
– hpaulj
Nov 19 '18 at 16:47
add a comment |
2
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
1
I don't understand where the2903comes from, andbis three levels deep while your for loop is only two. Also you're using the comparison==rather than the assignment=.
– Acccumulation
Nov 19 '18 at 16:34
If I gave you a functionupdate, would it have to updatebin place, so thatprint(b); update(b); print(b)would output two different values, or can it return a copynew_b = update(b). In the second case, do you need to retain the original object? That is, afternew_b = update(b), would you requireid(new_b) == id(b)? Would you require thatid(new_b[0]) == id(b[0])as well? (Keeping the existing rows, and only replacing the tuples within each row?)
– chepner
Nov 19 '18 at 16:38
ee== ..is not an assignment. Andee=..changes the variable but does not changen.
– hpaulj
Nov 19 '18 at 16:47
2
2
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
1
1
I don't understand where the
2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.– Acccumulation
Nov 19 '18 at 16:34
I don't understand where the
2903 comes from, and b is three levels deep while your for loop is only two. Also you're using the comparison == rather than the assignment =.– Acccumulation
Nov 19 '18 at 16:34
If I gave you a function
update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)– chepner
Nov 19 '18 at 16:38
If I gave you a function
update, would it have to update b in place, so that print(b); update(b); print(b) would output two different values, or can it return a copy new_b = update(b). In the second case, do you need to retain the original object? That is, after new_b = update(b), would you require id(new_b) == id(b)? Would you require that id(new_b[0]) == id(b[0]) as well? (Keeping the existing rows, and only replacing the tuples within each row?)– chepner
Nov 19 '18 at 16:38
ee== .. is not an assignment. And ee=.. changes the variable but does not change n.– hpaulj
Nov 19 '18 at 16:47
ee== .. is not an assignment. And ee=.. changes the variable but does not change n.– hpaulj
Nov 19 '18 at 16:47
add a comment |
2 Answers
2
active
oldest
votes
3
Tuples are immutable. You can use a list comprehension instead:
res = [[(i+5, j) for i, j in tup] for tup in b]
[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takesbas an argument and updates it in-place.
– chepner
Nov 19 '18 at 16:35
add a comment |
2
You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]
for n in b:
for i, (ee, ww) in enumerate(n):
n[i] = (ee + 2903, ww)
print(b)
Output:
[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
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',
autoActivateHeartbeat: false,
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
});
}
});
draft saved
draft discarded
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%2f53378909%2fhow-to-change-the-first-value-in-the-tuple-of-a-list%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
3
Tuples are immutable. You can use a list comprehension instead:
res = [[(i+5, j) for i, j in tup] for tup in b]
[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takesbas an argument and updates it in-place.
– chepner
Nov 19 '18 at 16:35
add a comment |
3
Tuples are immutable. You can use a list comprehension instead:
res = [[(i+5, j) for i, j in tup] for tup in b]
[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takesbas an argument and updates it in-place.
– chepner
Nov 19 '18 at 16:35
add a comment |
3
3
3
Tuples are immutable. You can use a list comprehension instead:
res = [[(i+5, j) for i, j in tup] for tup in b]
[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
Tuples are immutable. You can use a list comprehension instead:
res = [[(i+5, j) for i, j in tup] for tup in b]
[[(6, 0.044), (7, 0.042)], [(9, 0.18), (11, 0.023)], [(9, 0.03), (10, 0.023)]]
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
answered Nov 19 '18 at 16:29
jppjpp
100k2161111
100k2161111
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takesbas an argument and updates it in-place.
– chepner
Nov 19 '18 at 16:35
add a comment |
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takesbas an argument and updates it in-place.
– chepner
Nov 19 '18 at 16:35
3
3
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
This builds an entirely new matrix, rather than updating the existing one. Depending on the use case, this may not be useful.
– chepner
Nov 19 '18 at 16:32
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
@chepner, Sure. I think there are fundamental flaws in understanding for OP, e.g. the use of the term "matrix". I think we need more clarity to be sure a list comprehension does not suffice. Of course, time complexity is the same either way, there's no intrinsic benefit to modifying the "existing matrix" if either would work.
– jpp
Nov 19 '18 at 16:33
1
1
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes
b as an argument and updates it in-place.– chepner
Nov 19 '18 at 16:35
I read the original question ("... keep the change to the original matrix b") as indicating that this might be, for example, inside a function that takes
b as an argument and updates it in-place.– chepner
Nov 19 '18 at 16:35
add a comment |
2
You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]
for n in b:
for i, (ee, ww) in enumerate(n):
n[i] = (ee + 2903, ww)
print(b)
Output:
[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
add a comment |
2
You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]
for n in b:
for i, (ee, ww) in enumerate(n):
n[i] = (ee + 2903, ww)
print(b)
Output:
[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
add a comment |
2
2
2
You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]
for n in b:
for i, (ee, ww) in enumerate(n):
n[i] = (ee + 2903, ww)
print(b)
Output:
[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
You cannot modify tuples, they are immutable in Python. You can however replace the existing tuples with other tuples.
b = [[(1, 0.044), (2, 0.042)], [(4, 0.18), (6, 0.023)], [(4, 0.03), (5, 0.023)]]
for n in b:
for i, (ee, ww) in enumerate(n):
n[i] = (ee + 2903, ww)
print(b)
Output:
[[(2904, 0.044), (2905, 0.042)], [(2907, 0.18), (2909, 0.023)], [(2907, 0.03), (2908, 0.023)]]
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
answered Nov 19 '18 at 16:31
jdehesajdehesa
23.8k43554
23.8k43554
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f53378909%2fhow-to-change-the-first-value-in-the-tuple-of-a-list%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
2
Tuples are immutable; at the least, you need to replace each tuple with a new one in each sublist.
– chepner
Nov 19 '18 at 16:29
1
I don't understand where the
2903comes from, andbis three levels deep while your for loop is only two. Also you're using the comparison==rather than the assignment=.– Acccumulation
Nov 19 '18 at 16:34
If I gave you a function
update, would it have to updatebin place, so thatprint(b); update(b); print(b)would output two different values, or can it return a copynew_b = update(b). In the second case, do you need to retain the original object? That is, afternew_b = update(b), would you requireid(new_b) == id(b)? Would you require thatid(new_b[0]) == id(b[0])as well? (Keeping the existing rows, and only replacing the tuples within each row?)– chepner
Nov 19 '18 at 16:38
ee== ..is not an assignment. Andee=..changes the variable but does not changen.– hpaulj
Nov 19 '18 at 16:47