pandas dataframe count行值

Question

我有一个像下面这样的单词列表.

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

数据框如下所示.

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

输出:

    id path

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

我想计算路径数据以获得以下输出.是否有可能实现这种转变？

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

您可以使用矢量化字符串方法str.count()(请参阅文档和参考),并将wordlist中的每个元素提供给新数据帧:

In [4]: pd.DataFrame({name : df["path"].str.count(name) for name in wordlist})
Out[4]:
    p1  p2  p3  p4  p5  p6  p7
id
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

更新:评论的一些答案.实际上,如果字符串可以是彼此的子串,这将不起作用(但OP应该澄清它).如果是这种情况,这将起作用(并且也更快):

splitted = df["path"].str.split(",")
pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name in wordlist})

还有一些测试可以支持我更快的声明:-)
当然,我不知道实际用例是什么,但是我把数据框放大了一点(只重复了1000次,差异就大了) :

In [37]: %%timeit
   ....: splitted = df["path"].str.split(",")
   ....: pd.DataFrame({name : splitted.apply(lambda x: x.count(name)) for name i
n wordlist})
   ....:
100 loops, best of 3: 17.9 ms per loop

In [38]: %%timeit
   ....: pd.DataFrame({name:df["path"].str.count(name) for name in wordlist})
   ....:
10 loops, best of 3: 23.6 ms per loop

In [39]: %%timeit
   ....: c = df["path"].str.split(',').apply(Counter)
   ....: pd.DataFrame({n: c.apply(lambda x: x.get(n, 0)) for n in wordlist})
   ....:
10 loops, best of 3: 42.3 ms per loop

In [40]: %%timeit
   ....: dfN = df["path"].str.split(',').apply(lambda x: pd.Series(Counter(x)))
   ....: pd.DataFrame(dfN, columns=wordlist).fillna(0)
   ....:
1 loops, best of 3: 715 ms per loop

我也用更多的元素进行了测试wordlist,结论是:如果你有一个更大的数据帧,wordlist我的方法中元素数量相对较少,如果你有一个很大wordlist的方法,Counter来自@RomanPekar可以更快(但只有最后一个).

Answer 2

我认为这会很有效率

# 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

更新

另一种方法:

>>> 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

更新2

一些粗略的性能测试:

>>> 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

更新3

看完这个话题后我发现这Counter很慢.您可以使用defaultdict以下方法优化它:

>>> 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

和一些测试:

>>> 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