问题描述
所以基本上我想要 5 >np.nan 返回 np.nan 或 Nan 而不是 FALSE
在熊猫系列中，代码如下:
So basically I want 5 > np.nan return np.nan or Nan instead of FALSE
In pandas series, here's the code :
import pandas as pd
import numpy as np
a = pd.DataFrame({"x":[1,2,3,4],"y":[1,np.nan,5,1]})
a["x"]>a["y"]
将返回:
0 False
1 False
2 False
3 True
dtype: bool
我目前保存 Nan 信息的方法是:
My current approach to preserve the Nan information is :
value_comparison = a["x"]>a["y"]
nan_comparison = a["x"].isna() | a["y"].isna()
value_comparison.where(~nan_comparison,np.nan)
返回的地方
0 0.0
1 NaN
2 0.0
3 1.0
dtype: float64
我也采用了类似的方法进行 numpy 比较
I took the similar approach for numpy comparison too
即使结果是正确的，我相信我的解决方案不优雅，有没有更好的(pandas 和 numpy)方法来做到这一点，遵循 巨蟒之禅 ?(更好的可读性，更直接)
Even when the result is correct, I believe my solution is not elegant, is there any better(pandas and numpy) way to do this, which follows the zen of python ? (better readability, more straighforward)
推荐答案
仅稍微改进/(更改)您的解决方案:
Only a bit improved/(changed) your solution:
value_comparison = (a["x"]>a["y"])
nan_comparison = a[["x", "y"]].notna().all(axis=1)
#alternative
#nan_comparison = a["x"].notna() & a["y"].notna()
m = value_comparison.where(nan_comparison)
print (m)
0 0.0
1 NaN
2 0.0
3 1.0
dtype: float64
最后可以转换为可空布尔值:
Last is possible convert to nullable boolean:
m = value_comparison.where(nan_comparison).astype('boolean')
print (m)
0 False
1 <NA>
2 False
3 True
dtype: boolean
这篇关于在 Pandas 和 Numpy Python 中保留 Nan 的同时进行值比较的更优雅的方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！