Question

我有一个大熊猫系列，其元素构成了frozensets：

data = {0: frozenset({'apple', 'banana'}),
     1: frozenset({'apple', 'orange'}),
     2: frozenset({'banana'}),
     3: frozenset({'kumquat', 'orange'}),
     4: frozenset({'orange'}),
     5: frozenset({'orange', 'pear'}),
     6: frozenset({'orange', 'pear'}),
     7: frozenset({'apple', 'banana', 'pear'}),
     8: frozenset({'banana', 'persimmon'}),
     9: frozenset({'apple'}),
     10: frozenset({'banana'}),
     11: frozenset({'apple'})}

tokens = pd.Series(data); tokens

0           (apple, banana)
1           (orange, apple)
2                  (banana)
3         (orange, kumquat)
4                  (orange)
5            (orange, pear)
6            (orange, pear)
7     (apple, banana, pear)
8       (persimmon, banana)
9                   (apple)
10                 (banana)
11                  (apple)
Name: Tokens, dtype: object

我想成对应用一个函数。例如，tokens.diff给出了连续行之间的集合差异：

0                   NaN
1              (orange)
2              (banana)
3     (orange, kumquat)
4                    ()
5                (pear)
6                    ()
7       (apple, banana)
8           (persimmon)
9               (apple)
10             (banana)
11              (apple)
Name: Tokens, dtype: object

我想要同样的事情，但我想要在连续的行上设置联合而不是设置差异。所以，我理想地喜欢：

0                                 NaN
1             (orange, apple, banana)
2             (banana, orange, apply)
3           (orange, kumquat, banana)
4                   (orange, kumquat)
                                  ...

如何通过熊猫实现这一目标？我知道我可以使用zip和list comp来做到这一点，但希望有更好的方法。

Answer 1

几种方式

选项1] 列表理解

In [3631]: pd.Series([x[0].union(x[1])
                      for x in zip(tokens, tokens.shift(-1).fillna(''))],
                     index=tokens.index)
Out[3631]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

选项2] map

In [3632]: pd.Series(map(lambda x: x[0].union(x[1]), 
                         zip(tokens, tokens.shift(-1).fillna(''))),
                     index=tokens.index)
Out[3632]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

选项3] 使用concat和apply

In [3633]: pd.concat([tokens, tokens.shift(-1).fillna('')],
                     axis=1).apply(lambda x: x[0].union(x[1]), axis=1)
Out[3633]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

计时

In [3647]: tokens.shape
Out[3647]: (60000L,)

In [3648]: %timeit pd.Series([x[0].union(x[1]) for x in zip(tokens, tokens.shift(-1).fillna(''))], index=tokens.index)
10 loops, best of 3: 35 ms per loop

In [3649]: %timeit pd.Series(map(lambda x: x[0].union(x[1]), zip(tokens, tokens.shift(-1).fillna(''))), index=tokens.index)
10 loops, best of 3: 40.9 ms per loop

In [3650]: %timeit pd.concat([tokens, tokens.shift(-1).fillna('')], axis=1).apply(lambda x: x[0].union(x[1]), axis=1)
1 loop, best of 3: 2.2 s per loop

与diff

上的数字无关

In [3653]: %timeit tokens.diff()
10 loops, best of 3: 10.8 ms per loop

在pandas系列上成对应用函数

1 个答案: