在我的df文件中，下面的每個(gè)實(shí)體（Grubhub、Toasttab、Tenk）都有一列，并且每行的該列值中都顯示“是”或“否”，我有以下代碼，例如：df['Grubhub'] = df[['On GrubHub or Seamless?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['ToastTab'] = df[['On ToastTab?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['Tenk'] = df[['On Tenk?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['Udemy'] = df[['On Udmey?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['Postmates'] = df[['On Postmates?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['Doordash'] = df[['On DoorDash?']].apply(lambda x: any(x == 'Yes'), axis = 1)df['Google'] = df[['On Goole?']].apply(lambda x: any(x == 'Yes'), axis = 1)這為每個(gè)實(shí)體（ Grubhub、Toasttab、Tenk ）提供了一個(gè)新列，并且該列給出了 true 或 false 值，是否有一種更有效的方法可以在一行代碼或函數(shù)中完成所有這些操作？感謝您提前的幫助
1 回答
至尊寶的傳說(shuō)
TA貢獻(xiàn)1789條經(jīng)驗(yàn) 獲得超10個(gè)贊
您可以創(chuàng)建一個(gè)列圖并應(yīng)用function內(nèi)部loop：
columns_map = (
('Grubhub', 'On GrubHub or Seamless?'),
('ToastTab', 'On ToastTab?'),
('Tenk', 'On Tenk?'),
# etc ...
)
for new_col, alias in columns_map:
df[new_col] = df[alias].apply(lambda x: x == 'Yes')
# also you can easily remove aliases columns:
# df = df.drop(columns=[alias])
或者您可以將值設(shè)置到原始列中并根據(jù)需要重命名（不帶drop()）：
for new_col, alias in columns_map:
df[alias] = df[alias].apply(lambda x: x == 'Yes')
df.rename(
columns={alias: new_col for new_col, alias in columns_map},
inplace=True
)
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