我們有資料，如下所示：
import pandas as pd
import numpy as np
d = {'shift_city_id': [1, 1,2,3,3,3,1], 'closest_city_id': [np.nan, np.nan,np.nan,np.nan,np.nan,np.nan,np.nan]}
df = pd.DataFrame(data=d)
df
shift_city_id closest_city_id
0 1 NaN
1 1 NaN
2 2 NaN
3 3 NaN
4 3 NaN
5 3 NaN
6 1 NaN
目標是在具有特定 city_id 的單元格中逐一輸入特定值。所以我試試這個：
df.loc[lambda x: (x['shift_city_id']==3),'closest_city_id'].iloc[0]=3
df.loc[lambda x: (x['shift_city_id']==3),'closest_city_id'].iloc[1]=4
df.loc[lambda x: (x['shift_city_id']==3),'closest_city_id'].iloc[2]=5
并且在資料框中沒有任何變化，np.nan 仍然存在。結果應該是：
shift_city_id closest_city_id
0 1 NaN
1 1 NaN
2 2 NaN
3 3 3
4 3 4
5 3 5
6 1 NaN
什么可以解決問題？
uj5u.com熱心網友回復：
你可以這樣做：
df.loc[lambda x: (x['shift_city_id']==3),'closest_city_id'] = [3,4,5]
這給出了想要的輸出：
為了理解分配的不同行為，我遵循這些詳細的答案
uj5u.com熱心網友回復：
動態，以便您一次可以設定多個 id/值：
ids = [3]
id_values = {3: [3, 4, 5]}
df['closest_city_id'] = df[df['shift_city_id'].isin(ids)]['shift_city_id'].replace(id_values)
輸出：
shift_city_id closest_city_id
0 1 NaN
1 1 NaN
2 2 NaN
3 3 3.0
4 3 4.0
5 3 5.0
6 1 NaN
uj5u.com熱心網友回復：
這里 lambda 不是必需的，您可以比較列df['shift_city_id']，并且因為匹配的 3 行可以分配具有長度的串列3：
df.loc[df['shift_city_id']==3,'closest_city_id'] = [3,4,5]
print (df)
shift_city_id closest_city_id
0 1 NaN
1 1 NaN
2 2 NaN
3 3 3.0
4 3 4.0
5 3 5.0
6 1 NaN
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