a= {'A' : [1, 2,3,4],
'B' : ['FOOTBALL','BASKETBALL','HANDBALL','VOLLEYBALL'],
'C' : [[5,10,15,40],[1,4],[20,10,40],[10,40]]
}
如果 B 與 FOOTBALL 不同，我如何從 C 中洗掉元素 40 像這樣：
A B C
0 1 FOOTBALL [5, 10, 15, 40]
1 2 BASKETBALL [1, 4]
2 3 HANDBALL [20, 10]
3 4 VOLLEYBALL [10]
uj5u.com熱心網友回復：
您可以在掩碼中過濾兩邊，并40在串列中洗掉使用 lambda 函式：
df = pd.DataFrame(a)
m = df['B'].ne('FOOTBALL')
df.loc[m, 'C'] = df.loc[m, 'C'].apply(lambda x: [y for y in x if y!=40])
print (df)
A B C
0 1 FOOTBALL [5, 10, 15, 40]
1 2 BASKETBALL [1, 4]
2 3 HANDBALL [20, 10]
3 4 VOLLEYBALL [10]
選擇：
m = df['B'].ne('FOOTBALL')
df.loc[m, 'C'] = [[y for y in x if y!=40] for x in df.loc[m, 'C']]
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