如何在if语句中将一个值与"NaN"进行比较:Python 3.5.2 | Anaconda 4.1.0(64位)例如:a！= Nan其中a是浮点值 [英] How to compare one value with 'NaN' in if statement: Python 3.5.2 | Anaconda 4.1.0 (64 bit) For example: a!=Nan where a is either a float value

问题描述

for r in range(65000):
    for c in range(8):
        if df1.iloc[r,c] != NaN:
            k=k+1
            df.iloc[k,3] = df1.iloc[r,c]
        else:
            print("Nan Detected")
            l=l+1
print(l," Nan Values encountered")

推荐答案

不幸的是，NaN会比较false，甚至与它本身也一样.因此df1.iloc[r,c] != NaN始终为真.

Unfortunately NaN will compare false, even with itself. So df1.iloc[r,c] != NaN is always true.

使用numpy.isnan(number)或math.isnan(number)代替检查number是否为NaN.

Use numpy.isnan(number) or math.isnan(number) instead to check if number is NaN.

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