Python dictionary rename key and group
I am trying to rename some keys and group the values for the grouped keys. My content looks like this:
text_image_old = {10_pdf 10_pdf0: "some text", 10_pdf 10_pdf1: "more text", 10_pdf 10_pdf2: "even more text"}
Using regex, I can iteratively replace the names, such that only 10_pdf would be left, but due to the loop, the text would just contain the values "even more text" (e.g. the last value):
text_image_new = {re.sub('[a-zA-Z0-9_]+.pdf[0-9]', '', k): v for k, v in text_image_old.items()}
How could i replace the keys and group the values? Thank you!
Edit: the expected output should look like this
text_image_new = {10_pdf :"some text" "more text" "even more text"}
or if its easier to get:
text_image_new = {10_pdf :"some text more text even more text"}
python-3.x dictionary key
asked Nov 15 '18 at 14:54
SimonSimon
369
add a comment |
I am trying to rename some keys and group the values for the grouped keys. My content looks like this:
text_image_old = {10_pdf 10_pdf0: "some text", 10_pdf 10_pdf1: "more text", 10_pdf 10_pdf2: "even more text"}
Using regex, I can iteratively replace the names, such that only 10_pdf would be left, but due to the loop, the text would just contain the values "even more text" (e.g. the last value):
text_image_new = {re.sub('[a-zA-Z0-9_]+.pdf[0-9]', '', k): v for k, v in text_image_old.items()}
How could i replace the keys and group the values? Thank you!
Edit: the expected output should look like this
text_image_new = {10_pdf :"some text" "more text" "even more text"}
or if its easier to get:
text_image_new = {10_pdf :"some text more text even more text"}
python-3.x dictionary key
asked Nov 15 '18 at 14:54
SimonSimon
369
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02
add a comment |
I am trying to rename some keys and group the values for the grouped keys. My content looks like this:
text_image_old = {10_pdf 10_pdf0: "some text", 10_pdf 10_pdf1: "more text", 10_pdf 10_pdf2: "even more text"}
Using regex, I can iteratively replace the names, such that only 10_pdf would be left, but due to the loop, the text would just contain the values "even more text" (e.g. the last value):
text_image_new = {re.sub('[a-zA-Z0-9_]+.pdf[0-9]', '', k): v for k, v in text_image_old.items()}
How could i replace the keys and group the values? Thank you!
Edit: the expected output should look like this
text_image_new = {10_pdf :"some text" "more text" "even more text"}
or if its easier to get:
text_image_new = {10_pdf :"some text more text even more text"}
python-3.x dictionary key
asked Nov 15 '18 at 14:54
SimonSimon
369
I am trying to rename some keys and group the values for the grouped keys. My content looks like this:
text_image_old = {10_pdf 10_pdf0: "some text", 10_pdf 10_pdf1: "more text", 10_pdf 10_pdf2: "even more text"}
Using regex, I can iteratively replace the names, such that only 10_pdf would be left, but due to the loop, the text would just contain the values "even more text" (e.g. the last value):
text_image_new = {re.sub('[a-zA-Z0-9_]+.pdf[0-9]', '', k): v for k, v in text_image_old.items()}
How could i replace the keys and group the values? Thank you!
Edit: the expected output should look like this
text_image_new = {10_pdf :"some text" "more text" "even more text"}
or if its easier to get:
text_image_new = {10_pdf :"some text more text even more text"}
python-3.x dictionary key
python-3.x dictionary key
asked Nov 15 '18 at 14:54
SimonSimon
369
asked Nov 15 '18 at 14:54
SimonSimon
369
edited Nov 15 '18 at 16:02
Simon
asked Nov 15 '18 at 14:54
SimonSimon
369
asked Nov 15 '18 at 14:54
SimonSimon
369
asked Nov 15 '18 at 14:54
SimonSimon
369
369
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02
add a comment |
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02
add a comment |
1 Answer
1
active
oldest
votes
I hope this should work for you, or at least help solving your problem:
text_image_old = {'10_pdf 10_pdf0': "some text", '10_pdf 10_pdf1': "more text",
'10_pdf 10_pdf2': "even more text"}
new_dict = {}
for k, v in text_image_old.items():
k = k.split(' ')[0]
if k in new_dict:
new_dict[k] += v + ' '
else:
new_dict[k] = v + ' '
print(new_dict)
answered Nov 15 '18 at 16:30
DrakoDrako
501415
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
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
});
}
});
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%2f53322138%2fpython-dictionary-rename-key-and-group%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I hope this should work for you, or at least help solving your problem:
text_image_old = {'10_pdf 10_pdf0': "some text", '10_pdf 10_pdf1': "more text",
'10_pdf 10_pdf2': "even more text"}
new_dict = {}
for k, v in text_image_old.items():
k = k.split(' ')[0]
if k in new_dict:
new_dict[k] += v + ' '
else:
new_dict[k] = v + ' '
print(new_dict)
answered Nov 15 '18 at 16:30
DrakoDrako
501415
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
add a comment |
I hope this should work for you, or at least help solving your problem:
text_image_old = {'10_pdf 10_pdf0': "some text", '10_pdf 10_pdf1': "more text",
'10_pdf 10_pdf2': "even more text"}
new_dict = {}
for k, v in text_image_old.items():
k = k.split(' ')[0]
if k in new_dict:
new_dict[k] += v + ' '
else:
new_dict[k] = v + ' '
print(new_dict)
answered Nov 15 '18 at 16:30
DrakoDrako
501415
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
add a comment |
I hope this should work for you, or at least help solving your problem:
text_image_old = {'10_pdf 10_pdf0': "some text", '10_pdf 10_pdf1': "more text",
'10_pdf 10_pdf2': "even more text"}
new_dict = {}
for k, v in text_image_old.items():
k = k.split(' ')[0]
if k in new_dict:
new_dict[k] += v + ' '
else:
new_dict[k] = v + ' '
print(new_dict)
answered Nov 15 '18 at 16:30
DrakoDrako
501415
I hope this should work for you, or at least help solving your problem:
text_image_old = {'10_pdf 10_pdf0': "some text", '10_pdf 10_pdf1': "more text",
'10_pdf 10_pdf2': "even more text"}
new_dict = {}
for k, v in text_image_old.items():
k = k.split(' ')[0]
if k in new_dict:
new_dict[k] += v + ' '
else:
new_dict[k] = v + ' '
print(new_dict)
answered Nov 15 '18 at 16:30
DrakoDrako
501415
answered Nov 15 '18 at 16:30
DrakoDrako
501415
answered Nov 15 '18 at 16:30
DrakoDrako
501415
answered Nov 15 '18 at 16:30
DrakoDrako
501415
501415
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
add a comment |
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
you can still implement you regex here if necessary, I just use split because from what you provide in question it seems doing the job here
– Drako
Nov 15 '18 at 16:34
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.
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%2f53322138%2fpython-dictionary-rename-key-and-group%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
Can you please share your expected output?
– Mayank Porwal
Nov 15 '18 at 15:47
I have edited the question. Thank you for your comment
– Simon
Nov 15 '18 at 16:02