Python 基于条件将 Pandas DataFrame 列从 String 转换为 Int
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Convert Pandas DataFrame Column From String to Int Based on Conditional
提问by Adam_G
I have a dataframe that looks like
我有一个看起来像的数据框
df
df
viz a1_count a1_mean a1_std
n 3 2 0.816497
y 0 NaN NaN
n 2 51 50.000000
I want to convert the "viz" column to 0 and 1, based on a conditional. I've tried:
我想根据条件将“viz”列转换为 0 和 1。我试过了：
df['viz'] = 0 if df['viz'] == "n" else 1
but I get:
但我得到：
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
采纳答案by EdChum
You're trying to compare a scalar with the entire series which raise the ValueErroryou saw. A simple method would be to cast the boolean series to int:
您正在尝试将标量与提高ValueError您所看到的的整个系列进行比较。一个简单的方法是将布尔系列转换为int：
In [84]:
df['viz'] = (df['viz'] !='n').astype(int)
df
Out[84]:
viz a1_count a1_mean a1_std
0 0 3 2 0.816497
1 1 0 NaN NaN
2 0 2 51 50.000000
You can also use np.where:
您还可以使用np.where：
In [86]:
df['viz'] = np.where(df['viz'] == 'n', 0, 1)
df
Out[86]:
viz a1_count a1_mean a1_std
0 0 3 2 0.816497
1 1 0 NaN NaN
2 0 2 51 50.000000
Output from the boolean comparison:
布尔比较的输出：
In [89]:
df['viz'] !='n'
Out[89]:
0 False
1 True
2 False
Name: viz, dtype: bool
And then casting to int:
然后投射到int：
In [90]:
(df['viz'] !='n').astype(int)
Out[90]:
0 0
1 1
2 0
Name: viz, dtype: int32
