[英]Replacing a column value by another column value based on regex - Python
[英]Replacing a value in an column based on another Column
我正在尝试基于另一个现有列替换列中的值。
2列看起来像这样
id_30 DeviceInfoShort
Android SAMSUNG
iOS iOS
None Windows
None None
Mac MacOS
Windows Windows
None None
id_30列的“无”未在图片中显示。 我想要的是id_30列中的所有“ None”值,它将检查DeviceInfoShort中的值是否为“ Windows”,如果是,则将id_30中的“ None”替换为“ Windows”,否则替换为“ Android”
下面的代码是我所拥有的。 它工作正常,但运行了10分钟。 我想我可以在这里使用map / apply使其更快...是否有使用熊猫的更优雅的方式?
%%time
for r in train_all_data.index:
if train_all_data.loc[r, 'id_30'] == 'None':
if train_all_data.loc[r, 'DeviceInfoShort'] == 'Windows':
train_all_data.loc[r, 'id_30'] = 'Windows'
else:
train_all_data.loc[r, 'id_30'] = 'Android'
使用Pandas / Numpy where
:
df['id_30'] = df['id_30'].where(
df['id_30'].notna(),
np.where(df['DeviceInfoShort'] == 'Windows', 'Windows', 'Android'))
temp = train_all_data[train_all_data['id_30'] == 'None']
train_all_data.loc[temp, 'id_30'] = 'Andorid'
temp1 = train_all_data[(train_all_data['id_30'] == 'None') & (train_all_data['DeviceInfoShort'] == 'Windows')]
train_all_data.loc[temp1, 'id_30'] = 'Windows'
也许这会更快:
df['id_30'] = df.apply(lambda x: "Windows" if x.id_30 == "None" and x.DeviceInfoShort == "Windows" else "Android")
根据我的经验,使用apply()总是比循环遍历更快
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