Generating string from a dictionary

Question

Let's say I have something like this:

counts = {"chuck" : 1 , "annie" : 42, "jan" : 100}
lst = list(counts.keys())
lst.sort()
for key in lst:
    print(key, counts[key])

The output would be

chuck 1
annie 42
jan 100

If I wanted to achieve this effect with a string, the string would be represented as

"chuck 1\nannie 42\njan 100\n"

Now if I wanted to accomplish this string I could try something like:

for key, value in counts.items():
    if value == 1:
        counts[key] = "1\n"
    elif value == 42:
        counts[key] = "42\n"
    elif value == 100:
        counts[key] = "100\n"

temp = list(counts.items())
temp.sort()
myStr = str(temp)

for char in myStr:
    if char in "[(',)]":
        myStr = myStr.replace(char,'')

finalStr = "" + myStr[0:19] + "" + myStr[20:37] + "" + myStr[38:53] + ""

But yeah, I think I'm overcomplicating things... What would be a better and easy way?

Thanks!

Can you try to explain what you are after in human language, so we don't have to act as python interpreters to understand your question? — arekolek, Apr 01 '16 at 19:34
Do you want to be sorting by key or by value? The initial loop shows you sorting by key, but the output you show is sorted by value instead. — Blckknght, Apr 01 '16 at 19:42

idjaw · Answer 1 · 2016-04-01T20:12:02.373

2

You can achieve this by formatting each key, value as the string you expect per "{} {}".format(key, value)

Then, join on '\n':

"\n".join(["{} {}".format(key, value) for key, value in counts.items()])

Demo:

Input:

counts = {"chuck" : 1 , "annie" : 42, "jan" : 100}

Output:

jan 100
chuck 1
annie 42

Noticed that sorted was a requirement, so here it is with a sort:

"\n".join(["{} {}".format(key, value) for key, value in sorted(counts.items())])

output:

annie 42
chuck 1
jan 100

edited Apr 01 '16 at 20:12

answered Apr 01 '16 at 19:37

idjaw

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Great minds think alike ... at approximately the same speed, too. :) I believe that it is more efficient to use a list comprehension with `.join()` rather than a generator expression. – zondo Apr 01 '16 at 19:39
So far, just about every answer that uses a generator expression gets the comment. From what I've read, it's because `.join()` needs to know how much memory to allocate, so it converts the generator to a list anyway. Of course, it would be more efficient to skip the generator part and just go straight to the list. – zondo Apr 01 '16 at 19:43
Interesting. Thanks for the info. – idjaw Apr 01 '16 at 19:44
@zondo Indeed, on this particular example, `%timeit` says 2.4 µs vs. 3.24 µs in favor of list comprehension. I'm not sure if that's not premature optimization, though. BTW, using `*item` to unpack instead of `key, value`, seems to be marginally slower. – arekolek Apr 01 '16 at 19:52
Awesome. Thanks for checking that @arekolek – idjaw Apr 01 '16 at 19:52
@zondo Here is the answer to comprehension vs. expression. It's great. Thought I'd share: http://stackoverflow.com/a/9061024/1832539 – idjaw Apr 06 '16 at 03:43
@idjaw: Thanks. I'll be sure to use that link in the future. – zondo Apr 06 '16 at 10:30

arekolek · Answer 2 · 2016-04-01T20:21:30.133

1

Concise with sorting:

>>> '\n'.join(sorted([k+' '+str(v) for k, v in counts.items())])
'annie 42\nchuck 1\njan 100'

Sorting lines instead of items. It's actually faster this way as measured by %timeit.

edited Apr 01 '16 at 20:21

answered Apr 01 '16 at 20:03

arekolek

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score 0 · Answer 3 · answered Apr 01 '16 at 19:41

0

It's a bit clunky:

'\n'.join(["{0} {1}".format(a,b) for a,b in counts.items()])

answered Apr 01 '16 at 19:41

Josh English

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Jared Goguen · Answer 4 · 2016-04-01T20:12:38.430

0

If you're going for concise.

d = {"chuck" : 1 , "annie" : 42, "jan" : 100}

'\n'.join('{} {}'.format(*item) for item in sorted(d.items()))

edited Apr 01 '16 at 20:12

answered Apr 01 '16 at 20:01

Jared Goguen

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Maciej A. Czyzewski · Accepted Answer · 2016-04-01T19:46:46.223

-1

d = {"chuck" : 1 , "annie" : 42, "jan" : 100};
s = '\n'.join(['%s %s' % (key, value) for (key, value) in sorted(d.items(), key=lambda x: x[1])])
print(s) # 'chuck 1\nannie 42\njan 100'

d = {"chuck" : 1 , "annie" : 42, "jan" : 100};
s = '\n'.join(['%s %s' % (key, value) for (key, value) in sorted(d.items(), key=lambda x: x[0])])
print(s) # 'annie 42\nchuck 1\njan 100'

edited Apr 01 '16 at 19:46

answered Apr 01 '16 at 19:38

Maciej A. Czyzewski

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I really like this one. Thanks @Maciej A. Czyzewski – Riggs Apr 01 '16 at 19:50
@MaciejA.Czyzewski How did I not see that! :) OK. Thanks! (btw. I did not downvote you...) – idjaw Apr 01 '16 at 20:00
You don't need to use the list comprehension, a generator expression with work the same without the braces. Also, `operator.itemgetter` could be used instead of the `lambda` function. Finally, `format` is generally preferred over `%` substitution. – Jared Goguen Apr 01 '16 at 20:09

Generating string from a dictionary

5 Answers5