Question

我不确定是否可以轻松完成。

我有2个数据帧。在第一个（df1）中有一列有文本（'文本'），在第二个有2列，一列有一些排序文本（'subString'），第二列有一个得分（'得分'）。

当这些subString是第一个数据框中文本列的子字符串时，我想要的是总结第二个数据框中与subString字段关联的所有分数。

例如，如果我有这样的数据框：

df1 = pd.DataFrame({
        'ID':[1,2,3,4,5,6],
        'Texts':['this is a string',
                 'here we have another string',
                 'this one is completly different',
                 'one more',
                 'this is one more',
                 'and the last one'],
        'c':['C','C','C','C','C','C'],
        'd':['D','D','D','D','NaN','NaN']
    }, columns = ['ID','Texts','c','d'])
df1

Out[2]: 
   ID                            Texts  c    d
0   1                 this is a string  C    D
1   2      here we have another string  C    D
2   3  this one is completly different  C    D
3   4                         one more  C    D
4   5                 this is one more  C  NaN
5   6                 and the last one  C  NaN

还有另一个这样的数据框：

df2 = pd.DataFrame({
        'SubString':['This', 'one', 'this is', 'is one'],
        'Score':[0.5, 0.2, 0.75, -0.5]
    }, columns = ['SubString','Score'])
df2

Out[3]: 
  SubString  Score
0      This   0.50
1       one   0.20
2   this is   0.75
3    is one  -0.50

我想得到这样的东西：

df1['Score'] = 0.0
for index1, row1 in df1.iterrows():
    score = 0
    for index2, row2 in df2.iterrows():
        if row2['SubString'] in row1['Texts']:
            score += row2['Score']
    df1.set_value(index1, 'Score', score)
df1

Out[4]: 
   ID                            Texts  c    d  Score
0   1                 this is a string  C    D   0.75
1   2      here we have another string  C    D   0.00
2   3  this one is completly different  C    D  -0.30
3   4                         one more  C    D   0.20
4   5                 this is one more  C  NaN   0.45
5   6                 and the last one  C  NaN   0.20

是否有更少的乱码和更快的方法呢？

谢谢！

Answer 1

选项1

In [691]: np.array([np.where(df1.Texts.str.contains(x.SubString), x.Score, 0) 
                    for _, x in df2.iterrows()]
                   ).sum(axis=0)
Out[691]: array([ 0.75,  0.  , -0.3 ,  0.2 ,  0.45,  0.2 ])

选项2

In [674]: df1.Texts.apply(lambda x: df2.Score[df2.SubString.apply(lambda y: y in x)].sum())
Out[674]:
0    0.75
1    0.00
2   -0.30
3    0.20
4    0.45
5    0.20
Name: Texts, dtype: float64

注意：apply没有摆脱循环，它只是隐藏它们。

在给定列中查找多个字符串

1 个答案: