Lis*_*isa 5 python nan missing-data dummy-variable
如果包含缺失值,如何创建虚拟变量?我有以下数据,我想根据几个条件创建一个虚拟变量。我的问题是它会自动将我的缺失值转换为 0,但我想将它们保留为缺失值。
import pandas as pd
mydata = {'x' : [10, 50, np.nan, 32, 47, np.nan, 20, 5, 100, 62],
'y' : [10, 1, 5, np.nan, 47, np.nan, 8, 5, 100, 3]}
df = pd.DataFrame(mydata)
df["z"] = ((df["x"] >= 50) & (df["y"] <= 20)).astype(int)
print(df)
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创建布尔掩码时,您正在将整数与nans. 在您的情况下,df['x']=np.nan与 50 进行比较时,如果您将其转换为整数,您的掩码df['x'] >= 50将始终False等于。0您只需创建一个布尔掩码,该掩码等于包含列中的True所有行,然后分配给这些行。np.nan['x', 'y']np.nan
代码:
import pandas as pd
import numpy as np
mydata = {'x' : [10, 50, np.nan, 32, 47, np.nan, 20, 5, 100, 62],
'y' : [10, 1, 5, np.nan, 47, np.nan, 8, 5, 100, 3]}
df = pd.DataFrame(mydata)
df["z"] = ((df["x"] >= 50) & (df["y"] <= 20)).astype("uint32")
df.loc[df[["x", "y"]].isna().any(axis=1), "z"] = np.nan
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输出:
x y z
0 10.0 10.0 0.0
1 50.0 1.0 1.0
2 NaN 5.0 NaN
3 32.0 NaN NaN
4 47.0 47.0 0.0
5 NaN NaN NaN
6 20.0 8.0 0.0
7 5.0 5.0 0.0
8 100.0 100.0 0.0
9 62.0 3.0 1.0
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或者,如果您想要单行语句,则可以使用嵌套np.where语句:
df["z"] = np.where(
df.isnull().any(axis=1), np.nan, np.where((df["x"] >= 50) & (df["y"] <= 20), 1, 0)
)
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