字符串包含两个熊猫系列

Question

我有一个系列，在熊猫数据帧中有一些字符串。我想在相邻列中搜索该字符串的存在。

在下面的示例中，我想搜索"选择"系列中的字符串是否包含在"水果"系列中，在新列"choice_match"中返回真 (1( 或假 (0(。

示例数据帧：

import pandas as pd
d = {'ID': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'fruit': [
'apple, banana', 'apple', 'apple', 'pineapple', 'apple, pineapple',            'orange', 'apple, orange', 'orange', 'banana', 'apple, peach'],
'choice': ['orange', 'orange', 'apple', 'pineapple', 'apple', 'orange',  'orange', 'orange', 'banana', 'banana']}
df = pd.DataFrame(data=d)

所需数据帧：

import pandas as pd
d = {'ID': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 'fruit': [
'apple, banana', 'apple', 'apple', 'pineapple', 'apple, pineapple',   'orange', 'apple, orange', 'orange', 'banana', 'apple, peach'],
'choice': ['orange', 'orange', 'apple', 'pineapple', 'apple', 'orange',      'orange', 'orange', 'banana', 'banana'],
'choice_match': [0, 0, 1, 1, 1, 1, 1, 1, 1, 0]}
df = pd.DataFrame(data=d)

Answer 1

这是一种方法：

df['choice_match'] = df.apply(lambda row: row['choice'] in row['fruit'].split(','),
                              axis=1).astype(int)

解释

• df.apply axis=1循环遍历每一行并应用逻辑;它接受匿名lambda函数。
• row['fruit'].split(',')从fruit列创建一个列表。这是必要的，因此，例如，apple在pineapple中不考虑。
• 出于显示目的，需要astype(int)将布尔值转换为整数。

Answer 2

In [75]: df['choice_match'] = (df['fruit']
                                 .str.split(',s*', expand=True)
                                 .eq(df['choice'], axis=0)
                                 .any(1).astype(np.int8))
In [76]: df
Out[76]:
   ID     choice             fruit  choice_match
0   1     orange     apple, banana             0
1   2     orange             apple             0
2   3      apple             apple             1
3   4  pineapple         pineapple             1
4   5      apple  apple, pineapple             1
5   6     orange            orange             1
6   7     orange     apple, orange             1
7   8     orange            orange             1
8   9     banana            banana             1
9  10     banana      apple, peach             0

循序渐进：

In [78]: df['fruit'].str.split(',s*', expand=True)
Out[78]:
           0          1
0      apple     banana
1      apple       None
2      apple       None
3  pineapple       None
4      apple  pineapple
5     orange       None
6      apple     orange
7     orange       None
8     banana       None
9      apple      peach
In [79]: df['fruit'].str.split(',s*', expand=True).eq(df['choice'], axis=0)
Out[79]:
       0      1
0  False  False
1  False  False
2   True  False
3   True  False
4   True  False
5   True  False
6  False   True
7   True  False
8   True  False
9  False  False
In [80]: df['fruit'].str.split(',s*', expand=True).eq(df['choice'], axis=0).any(1)
Out[80]:
0    False
1    False
2     True
3     True
4     True
5     True
6     True
7     True
8     True
9    False
dtype: bool
In [81]: df['fruit'].str.split(',s*', expand=True).eq(df['choice'], axis=0).any(1).astype(np.int8)
Out[81]:
0    0
1    0
2    1
3    1
4    1
5    1
6    1
7    1
8    1
9    0
dtype: int8

Answer 3

选项 1
使用Numpy的find
当find找不到该值时，它将返回-1

from numpy.core.defchararray import find
choice = df.choice.values.astype(str)
fruit = df.fruit.values.astype(str)
df.assign(choice_match=(find(fruit, choice) > -1).astype(np.uint))
   ID     choice             fruit  choice_match
0   1     orange     apple, banana             0
1   2     orange             apple             0
2   3      apple             apple             1
3   4  pineapple         pineapple             1
4   5      apple  apple, pineapple             1
5   6     orange            orange             1
6   7     orange     apple, orange             1
7   8     orange            orange             1
8   9     banana            banana             1
9  10     banana      apple, peach             0

---

选项 2
设置逻辑
对于set，<是严格子集，<=是子集。 让自己set pd.Series一些，并使用<=来确定一列的集合是否是另一列集合的子集。

choice = df.choice.apply(lambda x: set([x]))
fruit = df.fruit.str.split(', ').apply(set)
df.assign(choice_match=(choice <= fruit).astype(np.uint))
   ID     choice             fruit  choice_match
0   1     orange     apple, banana             0
1   2     orange             apple             0
2   3      apple             apple             1
3   4  pineapple         pineapple             1
4   5      apple  apple, pineapple             1
5   6     orange            orange             1
6   7     orange     apple, orange             1
7   8     orange            orange             1
8   9     banana            banana             1
9  10     banana      apple, peach             0

---

选项 3
灵感来自@Wen的回答
使用get_dummies和max

c = pd.get_dummies(df.choice)
f = df.fruit.str.get_dummies(', ')
df.assign(choice_match=pd.DataFrame.mul(*c.align(f, 'inner')).max(1))
   ID     choice             fruit  choice_match
0   1     orange     apple, banana             0
1   2     orange             apple             0
2   3      apple             apple             1
3   4  pineapple         pineapple             1
4   5      apple  apple, pineapple             1
5   6     orange            orange             1
6   7     orange     apple, orange             1
7   8     orange            orange             1
8   9     banana            banana             1
9  10     banana      apple, peach             0

Answer 4

um找到一个有趣的方法get_dummies

(df.fruit.str.replace(' ','').str.get_dummies(',')+df.choice.str.get_dummies()).gt(1).any(1)
Out[726]: 
0    False
1    False
2     True
3     True
4     True
5     True
6     True
7     True
8     True
9    False
dtype: bool

重新分配后

df['New']=(df.fruit.str.replace(' ','').str.get_dummies(',')+df.choice.str.get_dummies()).gt(1).any(1).astype(int)
df
Out[728]: 
   ID     choice             fruit  New
0   1     orange     apple, banana    0
1   2     orange             apple    0
2   3      apple             apple    1
3   4  pineapple         pineapple    1
4   5      apple  apple, pineapple    1
5   6     orange            orange    1
6   7     orange     apple, orange    1
7   8     orange            orange    1
8   9     banana            banana    1
9  10     banana      apple, peach    0