KeyError: False當我運行這條線時，我得到了：
df['Eligible'] = df[('DeliveryOnTime' == "On-time") | ('DeliveryOnTime' == "Early")]
我一直在試圖找到一種方法來使用np.whereand來執行這個條件，.loc()但都沒有奏效。對如何Eligible使用來自的資料將條件應用于新列的其他想法持開放態度DeliveryOnTime
我試過這些：
np.where
df['Eligible'] = np.where((df['DeliveryOnTime'] == "On-time") | (df['DeliveryOnTime'] == "Early"), 1, 1)
.loc()
df['Eligible'] = df.loc[(df['DeliveryOnTime'] == "On-time") & (df['DeliveryOnTime'] == "Early"), 'Total Orders'].sum()
樣本資料：
data = {'ID': [1, 1, 1, 2, 2, 3, 4, 5, 5],
'DeliveryOnTime': ["On-time", "Late", "Early", "On-time", "On-time", "Late", "Early", "Early", "On-time"],
}
df = pd.DataFrame(data)
#For the sake of example data, the count of `DeliveryOnTime` will be the total number of orders.
df['Total Orders'] = df['DeliveryOnTime'].count()
uj5u.com熱心網友回復：
正確的語法是：
df['Eligible'] = (df['DeliveryOnTime'] == "On-time") | (df['DeliveryOnTime'] == "Early")
# OR
df['Eligible'] = df['DeliveryOnTime'].isin(["On-time", "Early"])
輸出：
>>> df
ID DeliveryOnTime Total Orders Eligible
0 1 On-time 9 True
1 1 Late 9 False
2 1 Early 9 True
3 2 On-time 9 True
4 2 On-time 9 True
5 3 Late 9 False
6 4 Early 9 True
7 5 Early 9 True
8 5 On-time 9 True
uj5u.com熱心網友回復：
該df參考錯了地方。請嘗試：
df['Elegible'] = (df['DeliveryOnTime'] == "On-time") | (df['DeliveryOnTime'] =="Early")
輸出：
>>> df
ID DeliveryOnTime Elegible
0 1 On-time True
1 1 Late False
2 1 Early True
3 2 On-time True
4 2 On-time True
5 3 Late False
6 4 Early True
7 5 Early True
8 5 Early True
uj5u.com熱心網友回復：
您不能直接呼叫這些列。查看檢測所有“準時”或“提前”行的解決方案
df["eligible"] = df.DeliveryOnTime.isin(['On-time', 'Early'])
df['eligible'].groupby(df['ID']).transform('sum')
df
ID DeliveryOnTime eligible TotalOrders
0 1 On-time True 2
1 1 Late False 2
2 1 Early True 2
3 2 On-time True 2
4 2 On-time True 2
5 3 Late False 0
6 4 Early True 1
7 5 Early True 1
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