Counting equal strings in Python

Question

I have a list of strings and some of them are equal. I need some script which would count equal strings. Ex:

I have a list with some words :

"House"
"Dream"
"Tree"
"Tree"
"House"
"Sky"
"House"

And the output should look like this:

"House" - 3
"Tree" - 2
"Dream" - 1
and so on

sort file.txt | uniq -c will do what you want on unix or cygwin. Otherwise, if this is an assignment, you need to tell us what you tried already and what didn't work about it. — Karl Bielefeldt
– Karl Bielefeldt, Commented Dec 2, 2011 at 23:41

Raymond Hettinger · Accepted Answer · 2011-12-02 23:39:43Z

8

Use collections.Counter(). It is designed for exactly this use case:

>>> import collections
>>> seq = ["House", "Dream", "Tree", "Tree", "House", "Sky", "House"]
>>> for word, cnt in collections.Counter(seq).most_common():
        print repr(word), '-', cnt

'House' - 3
'Tree' - 2
'Sky' - 1
'Dream' - 1

answered Dec 2, 2011 at 23:39

Raymond Hettinger

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2 Comments

TJD Over a year ago

This is a great solution, but note that Counter only exists in Python 2.7+

Raymond Hettinger Over a year ago

There is a Py2.5 and Py2.6 backport of Counter at code.activestate.com/recipes/576611

Tadeck · Accepted Answer · 2011-12-02 23:55:44Z

5

Solution

This is quite simple (words is a list of words you want to process):

result = {}
for word in set(words):
    result[word] = words.count(word)

It does not require any additional modules.

Test

For the following words value:

words = ['House', 'Dream', 'Tree', 'Tree', 'House', 'Sky', 'House']

it will give you the following result:

>>> result
{'Dream': 1, 'House': 3, 'Sky': 1, 'Tree': 2}

Does it answer your question?

answered Dec 2, 2011 at 23:55

Tadeck

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3 Comments

Raymond Hettinger Over a year ago

If you want to avoid the standard library for some reason, it would be better to replace words.count(word) with result.get(word, 0) + 1. This simple change replaces an O(n) operation with an O(1) operation.

Tadeck Over a year ago

@RaymondHettinger: The change maybe is simple, but more complex than you proposed. At least your proposal does not work. Did you want to say that I should replace set(words) with words and result[word] = words.count(word) with result[word] = result.get(word, 0) + 1?

Raymond Hettinger Over a year ago

result = {} and for word in words: result[word] = result.get(word, 0) + 1 and if you want to put a bow-tie on it: for word, cnt in sorted(result.items(), reverse=True): print repr(word), '-', cnt

sverre · Accepted Answer · 2011-12-03 02:56:30Z

3

from collections import defaultdict
counts = defaultdict(int)
for s in strings:
    counts[s] += 1
for (k, v) in counts.items():
    print '"%s" - %d' % (k, v)

edited Dec 3, 2011 at 2:56

answered Dec 2, 2011 at 23:41

sverre

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1 Comment

joaquin Over a year ago

defaultdict(int) is enough. You don't need lambdas here

yasith · Accepted Answer · 2011-12-03 00:01:58Z

2

I will extend Tadeck's answer to print the results.

for word in set(words):
  print '''"%s" - %d''' %(word, words.count(word))

answered Dec 3, 2011 at 0:01

yasith

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2 Comments

Raymond Hettinger Over a year ago

The same comment applies as with Tadeck's solution. Using words.count(word) is an O(n) solution. You're much better-off using a python dictionary with its O(1) lookups.

Tadeck Over a year ago

@RaymondHettinger: Same comment as below my answer: your comment is about word in words loop, not word in set(words) loop, correct?

Kannaiyan · Accepted Answer · 2011-12-03 00:21:37Z

1

Below code should get you as expected

stringvalues = ['House', 'Home', 'House', 'House', 'Home']
for str in stringvalues:
    if( str in newdict ):
        newdict[str] = newdict[str] + 1
    else:
        newdict[str] = 1
all = newdict.items()
for k,v in all:
    print "%s-%s" % (k,v)

answered Dec 3, 2011 at 0:21

Kannaiyan

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