如何從其他列值有條件地創建 pandas 列

Question

我有一個看起來像這樣的 dataframe：

    word              start  stop speaker
0  but that's alright  2.72  3.47  2
1       we'll have to  8.43  9.07  1
2           okay sure  9.19 11.01  2
3               what? 11.02 12.00  1
4             I agree 12.01 14.00  2
5      but i disagree 14.01 17.00  2
6          thats fine 17.01 19.00  1
7     however you are 19.01 22.00  1
8           like this 22.01 24.00  1
9                 and 24.01 25.00  1

我想創建兩個新列，df.speaker_1 和 df.speaker_2。 當 df.speaker == 2 時，我希望 df.speaker_2 包含 df.word 的值。 當 df.speaker,= 2. 我希望它包含一個空字符串。 對其他揚聲器值重復相同的操作：它應該如下所示：

    word        start  stop speaker   speaker_2            speaker_1
0  but that's alright  2.72  3.47  2  but that's alright   
1       we'll have to  8.43  9.07  1                       we'll have to
2           okay sure  9.19 11.01  2  okay sure
3               what? 11.02 12.00  1                       what?
4             I agree 12.01 14.00  2  I agree
5      but i disagree 14.01 17.00  2  but i disagree
6          thats fine 17.01 19.00  1                       thats fine
7     however you are 19.01 22.00  1                       however you are
8           like this 22.01 24.00  1                       like this
9                 and 24.01 25.00  1                       and

任何建議將不勝感激，謝謝。

Answer 1

您可以從列word中復制值，然后根據需要替換為空字符串：

df['speaker_1'] = df['word']
df['speaker_2'] = df['word']

df.loc[df['speaker'] != 1, 'speaker_1'] = ''
df.loc[df['speaker'] != 2, 'speaker_2'] = ''

或者，您可以使用apply ，但我發現在您的情況下這更簡單。

Answer 2

您可以使用pd.DataFrame.mask() ：

df['speaker_1'] = df.word.mask(df.speaker!=1, '')
df['speaker_2'] = df.word.mask(df.speaker!=2, '')

#                  word  start  ...        speaker_1           speaker_2
# 0  but that's alright   2.72  ...                   but that's alright
# 1       we'll have to   8.43  ...    we'll have to                    
# 2           okay sure   9.19  ...                            okay sure
# 3               what?  11.02  ...            what?                    
# 4             I agree  12.01  ...                              I agree
# 5      but i disagree  14.01  ...                       but i disagree
# 6          thats fine  17.01  ...       thats fine                    
# 7     however you are  19.01  ...  however you are                    
# 8           like this  22.01  ...        like this                    
# 9                 and  24.01  ...              and