从熊猫数据框中替换单元格值

Question

这是示例数据。每当在Not win列中看到Not fail或Event时，我想将Time列的第1行的值更改为0。

x = pd.DataFrame({'Id': [102,303,944, 111, 675, 236], 'Event':['win', 'fail','Not fail','win','win','Not win'],
               'Time':[10,22,0, 23, 45, 0]})

这就是结果的样子。

        Event       Id   Time
0       win         102  10
1       fail        303  0
2       Not fail    944  0
3       win         111  23
4       win         675  0
5       Not win     236  0

Answer 1

尝试一下

l=['Not win','Not fail']
mask=df['Event'].isin(l).shift(-1).fillna(False)
df.loc[mask,'Time']=0

输出：

      Event   Id  Time
0       win  102    10
1      fail  303     0
2  Not fail  944     0
3       win  111    23
4       win  675     0
5   Not win  236     0

Answer 2

x["isPreviousEventNotFailNotWin"] = ((x.Event == "Not fail") | (x.Event == "Not win")).shift(-1)
x.Time[x.isPreviousEventNotFailNotWin == True] = 0

Answer 3

您可以通过以下方式使用where：

l = ['Not fail', 'Not win']
x['Time'] = np.where(x['Event'].shift(-1).isin(l), 0, x['Time'])

输出：

      Event   Id  Time
0       win  102    10
1      fail  303     0
2  Not fail  944     0
3       win  111    23
4       win  675     0
5   Not win  236     0

Answer 4

使用带有pd.Series.shift，后跟pd.DataFrame.loc的布尔掩码：

mask = x['Event'].isin(['Not fail', 'Not win']).shift(-1).fillna(False)

x.loc[mask, 'Time'] = 0

print(x)

      Event   Id  Time
0       win  102    10
1      fail  303     0
2  Not fail  944     0
3       win  111    23
4       win  675     0
5   Not win  236     0