pandas dataframe count row values

Question

I have a word list like following.

wordlist = ['p1','p2','p3','p4','p5','p6','p7']

And the dataframe is like following.

df = pd.DataFrame({'id' : [1,2,3,4],
                'path'  : ["p1,p2,p3,p4","p1,p2,p1","p1,p5,p5,p7","p1,p2,p3,p3"]})

output:

    id path

    1 p1,p2,p3,p4
    2 p1,p2,p1
    3 p1,p5,p5,p7
    4 p1,p2,p3,p3

I want to count the path data to get following output. Is it possible to get this kind of transformation?

id p1 p2 p3 p4 p5 p6 p7
1  1  1  1  1  0  0  0
2  2  1  0  0  0  0  0
3  1  0  0  0  2  0  1
4  1  1  2  0  0  0  0

Answer 1

I think this would be efficient

# create Series with dictionaries
>>> from collections import Counter
>>> c = df["path"].str.split(',').apply(Counter)
>>> c
0    {u'p2': 1, u'p3': 1, u'p1': 1, u'p4': 1}
1                        {u'p2': 1, u'p1': 2}
2              {u'p1': 1, u'p7': 1, u'p5': 2}
3              {u'p2': 1, u'p3': 2, u'p1': 1}

# create DataFrame
>>> pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})
   p1  p2  p3  p4  p5  p6  p7
0   1   1   1   1   0   0   0
1   2   1   0   0   0   0   0
2   1   0   0   0   2   0   1
3   1   1   2   0   0   0   0

update

Another way to do this:

>>> dfN = df["path"].str.split(',').apply(lambda x: pd.Series(Counter(x)))
>>> pd.DataFrame(dfN, columns=wordlist).fillna(0)
   p1  p2  p3  p4  p5  p6  p7
0   1   1   1   1   0   0   0
1   2   1   0   0   0   0   0
2   1   0   0   0   2   0   1
3   1   1   2   0   0   0   0

update 2

Some rough tests for performance:

>>> dfL = pd.concat([df]*100)
>>> timeit('c = dfL["path"].str.split(",").apply(Counter); d = pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd; from collections import Counter', number=100)
0.7363274283027295

>>> timeit('splitted = dfL["path"].str.split(","); d = pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd', number=100)
0.5305424618886718

# now let's make wordlist larger
>>> wordlist = wordlist + list(lowercase) + list(uppercase)

>>> timeit('c = dfL["path"].str.split(",").apply(Counter); d = pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd; from collections import Counter', number=100)
1.765344003293876

>>> timeit('splitted = dfL["path"].str.split(","); d = pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name in wordlist})', 'from __main__ import dfL, wordlist; import pandas as pd', number=100)
2.33328927599905

update 3

after reading this topic I've found that Counter is really slow. You can optimize it a bit by using defaultdict:

>>> def create_dict(x):
...     d = defaultdict(int)
...     for c in x:
...         d[c] += 1
...     return d
>>> c = df["path"].str.split(",").apply(create_dict)
>>> pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})
   p1  p2  p3  p4  p5  p6  p7
0   1   1   1   1   0   0   0
1   2   1   0   0   0   0   0
2   1   0   0   0   2   0   1
3   1   1   2   0   0   0   0

and some tests:

>>> timeit('c = dfL["path"].str.split(",").apply(create_dict); d = pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})', 'from __main__ import dfL, wordlist, create_dict; import pandas as pd; from collections import defaultdict', number=100)
0.45942801555111146

# now let's make wordlist larger
>>> wordlist = wordlist + list(lowercase) + list(uppercase)
>>> timeit('c = dfL["path"].str.split(",").apply(create_dict); d = pd.DataFrame({n: c.apply(lambda x: x[n]) for n in wordlist})', 'from __main__ import dfL, wordlist, create_dict; import pandas as pd; from collections import defaultdict', number=100)
1.5798653213942089