python - 如何单独获取匹配的ID作为新列？

Question

我有以下数据。

df['íd'] = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O']
df['value'] = ['12','12','13','12','13','12','13','12','12','15','17','12','13','12','15']

如果id对应的值等于12，如何单独获取id作为新列？

Answer 1

我认为你需要loc或where :

df = pd.DataFrame()
df['id'] = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O']
df['value'] = ['12','12','13','12','13','12','13','12','12','15','17','12','13','12','15']

df.loc[df['value'] == '12', 'new'] = df['id']
print (df)
   id value  new
0   A    12    A
1   B    12    B
2   C    13  NaN
3   D    12    D
4   E    13  NaN
5   F    12    F
6   G    13  NaN
7   H    12    H
8   I    12    I
9   J    15  NaN
10  K    17  NaN
11  L    12    L
12  M    13  NaN
13  N    12    N
14  O    15  NaN

<小时/>

df['new'] = df['id'].where(df['value'] == '12')
print (df)
   id value  new
0   A    12    A
1   B    12    B
2   C    13  NaN
3   D    12    D
4   E    13  NaN
5   F    12    F
6   G    13  NaN
7   H    12    H
8   I    12    I
9   J    15  NaN
10  K    17  NaN
11  L    12    L
12  M    13  NaN
13  N    12    N
14  O    15  NaN

编辑:

为了与数字进行比较，请删除'':

df = pd.DataFrame()
df['id'] = ['A','B','C','D','E','F','G','H','I','J','K','L','M','N','O']
df['value'] = [12,12,13,12,13,12,13,12,12,15,17,12,13,12,15]

df.loc[df['value'] == 12, 'new'] = df['id']
print (df)
   id  value  new
0   A     12    A
1   B     12    B
2   C     13  NaN
3   D     12    D
4   E     13  NaN
5   F     12    F
6   G     13  NaN
7   H     12    H
8   I     12    I
9   J     15  NaN
10  K     17  NaN
11  L     12    L
12  M     13  NaN
13  N     12    N
14  O     15  NaN