计算pandas中数据框中每列中值的更改,忽略NaN更改

时间:2017-07-11 14:54:53

标签: python pandas dataframe

我正在尝试计算pandas中数据框中每列中值的更改次数。除了NaN之外,我的代码工作得很好:如果一个列包含两个后续的NaN,它将被视为值的变化,这是我不想要的。我怎么能避免这种情况?

我做如下(感谢unutbu's answer):

import pandas as pd
import numpy as np

frame = pd.DataFrame({
    'time':[1234567000 , np.NaN, np.NaN],
    'X1':[96.32,96.01,96.05],
    'X2':[23.88,23.96,23.96]
},columns=['time','X1','X2']) 

print(frame)

changes = (frame.diff(axis=0) != 0).sum(axis=0)
print(changes)

changes = (frame != frame.shift(axis=0)).sum(axis=0)
print(changes)

返回:

           time     X1     X2
0  1.234567e+09  96.32  23.88
1           NaN  96.01  23.96
2           NaN  96.05  23.96

time    3
X1      3
X2      2
dtype: int64

time    3
X1      3
X2      2
dtype: int64

相反,结果应该是(注意时间列的变化):

time    2
X1      3
X2      2
dtype: int64

1 个答案:

答案 0 :(得分:3)

change = (frame.fillna(0).diff() != 0).sum()

输出:

time    2
X1      3
X2      2
dtype: int64

NaN是"truthy"。将NaN更改为零然后评估。

nan - nan = nan

nan != 0  = True

fillna(0)

0 - 0 = 0

0 != 0 = False