Pandas:查找字符串中子字符串的开始和结束位置

Question

from numpy.core.defchararray import find
df = pd.DataFrame({
"string": ["abc", "def", "ghi"],
"substring": ["bc", "e", "ghi"]
})

我得到了以下来确定开始位置，但我不确定如何获得结束位置：

df.assign(start=find(df['string'].values.astype(str),df['substring'].values.astype(str)))

预期结果：

string substring start end
abc    bc        1     2
def    e         1     1
ghi    ghi       0     2

Answer 1

将列表理解与:=一起用于元组中end字符串值的表达式中的变量赋值，最后一次赋值到新列：

df[['start','end']]=[(c:=a.find(b),c+len(b)-1) for a,b in zip(df['string'],df['substring'])]
print (df)
string substring  start  end
0    abc        bc      1    2
1    def         e      1    1
2    ghi       ghi      0    2

您的解决方案应该按照相同的逻辑进行更改：

from numpy.core.defchararray import find
df=df.assign(start=find(df['string'].values.astype(str),df['substring'].values.astype(str)),
end = lambda x: x['start'] + x['substring'].str.len() - 1)
print (df)
string substring  start  end
0    abc        bc      1    2
1    def         e      1    1
2    ghi       ghi      0    2

如果没有匹配返回-1，那么可能的解决方案应该在下一步中设置为NaN：

df = pd.DataFrame({
"string": ["ab7c", "def", "ghi"],
"substring": ["bc", "e", "ghi"]
})
print (df)
string substring
0   ab7c        bc
1    def         e
2    ghi       ghi
from numpy.core.defchararray import find
df=df.assign(start=find(df['string'].values.astype(str),df['substring'].values.astype(str)),
end = lambda x: x['start'] + x['substring'].str.len() - 1)
df[['start','end']] = df[['start','end']].mask(df['start'].eq(-1))
print (df)
string substring  start  end
0   ab7c        bc    NaN  NaN
1    def         e    1.0  1.0
2    ghi       ghi    0.0  2.0

Answer 2

另一种具有更好代码可读性的方法可以如下

## this will ensure if not found it will return None
def index_of_substring(main_string, substring):
try:
start_index = main_string.index(substring)
end_index = start_index + len(substring) -1
return(pd.Series([start_index,end_index]))
except ValueError:
return(pd.Series([None,None]))
## Then you call the function as follows
df = pd.DataFrame({
"string": ["abc", "def", "ghi"],
"substring": ["bc", "e", "ghi"]
})
df[["start","end"]] = df.apply(lambda row:index_of_substring(row['string'],row["substring"]),axis=1)
df.head()
string  substring   start   end
0   abc bc  1   2
1   def e   1   1
2   ghi ghi 0   2