python dictionary rotation result
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 at 18:30
luana nastasi
113
add a comment |
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 at 18:30
luana nastasi
113
add a comment |
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
asked Nov 22 at 18:30
luana nastasi
113
i have this dictionary:
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
if I do the print I have this output:
Surname White
Red
Name Mary
Bob
but I would like this output
Name Surname
Mary White
Bob Red
how can I do it without using pandas?
python
python
asked Nov 22 at 18:30
luana nastasi
113
asked Nov 22 at 18:30
luana nastasi
113
asked Nov 22 at 18:30
luana nastasi
113
asked Nov 22 at 18:30
luana nastasi
113
asked Nov 22 at 18:30
luana nastasi
113
113
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 at 20:08
planetmaker
4,50421629
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 at 18:55
ggorlen
6,3963825
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%2f53436486%2fpython-dictionary-rotation-result%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
add a comment |
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
add a comment |
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
Here's code that will sort keys of your dict (so you will get same out every time for same data) and print your dict in format you supplied with any amounts of keys and values of your dict.
dict1 = {'Name': ['Mary','Bob'],'Surname': ['White','Red']}
sorted_keys = sorted(dict1.keys())
spacing = 10
print_sheme = ('{:<' + str(spacing) + '}')*len(list(dict1.values())[0])
print(print_sheme.format(*sorted_keys))
for i in range(len(list(dict1.values())[0])):
values_to_print = [dict1[key][i] for key in sorted_keys]
print(print_sheme.format(*values_to_print))
Output:
Name Surname
Mary White
Bob Red
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
answered Nov 22 at 18:46
Filip Młynarski
1,5401311
1,5401311
add a comment |
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 at 20:08
planetmaker
4,50421629
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 at 20:08
planetmaker
4,50421629
add a comment |
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 at 20:08
planetmaker
4,50421629
A less general, but much shorter answer. Just iterate over both lists (but you can extend this to as how many lists you need to reference simultaneously):
print("{:10} {:10}".format("Name", "Surname"))
for (name, surname) in zip(dict1.get('Name'), dict1.get('Surname')):
print("{:10} {:10}".format(name, surname))
answered Nov 22 at 20:08
planetmaker
4,50421629
edited Nov 22 at 20:18
answered Nov 22 at 20:08
planetmaker
4,50421629
answered Nov 22 at 20:08
planetmaker
4,50421629
answered Nov 22 at 20:08
planetmaker
4,50421629
4,50421629
add a comment |
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 at 18:55
ggorlen
6,3963825
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 at 18:55
ggorlen
6,3963825
add a comment |
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 at 18:55
ggorlen
6,3963825
Here's a pretty straightforward implementation that handles arbitrary keys, uneven lists, and dynamically-sized columns:
def column_print(d, spacing=2):
columns = [[x] + y for x, y in zip(d.keys(), d.values())]
col_widths = [spacing + max([len(x) for x in e]) for e in columns]
for i in range(max([len(x) for x in columns])):
for w, col in zip(col_widths, columns):
print((col[i] if i < len(col) else "").ljust(w), end="")
print()
column_print({'Name': ['Mary','Bob','Mark'], 'Surname': ['White','Red','Blue','Black'], 'MI': ['S','A']})
Output:
Name Surname MI
Mary White S
Bob Red A
Mark Blue
Black
Try it!
answered Nov 22 at 18:55
ggorlen
6,3963825
edited Nov 22 at 20:35
answered Nov 22 at 18:55
ggorlen
6,3963825
answered Nov 22 at 18:55
ggorlen
6,3963825
answered Nov 22 at 18:55
ggorlen
6,3963825
6,3963825
add a comment |
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%2f53436486%2fpython-dictionary-rotation-result%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
