pandas :浮点数的差异 [英] Pandas: Difference of float numbers

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问题描述

我在这里遇到了一个奇怪的问题.我有一个数据框df,如下所示:

I've ran into a weird issue here. I have a dataframe df like below:

In [1561]: df
Out[1561]: 
      A     B
0  16.3  1.10
1  23.2  1.33
2  10.7 -0.43
3   5.7 -2.01
4   5.4 -1.86
5  23.5  3.14

我正在比较列A的每两个相邻行,并将差异存储在新列中:

I'm comparing every two adjacent rows of column A and storing the difference in a new column:

In [1562]: df['new_diff'] = (df.A - df.A.shift(-1)).fillna(0)
In [1563]: df
Out[1563]: 
      A     B  new_diff
0  16.3  1.10      -6.9
1  23.2  1.33      12.5
2  10.7 -0.43       5.0
3   5.7 -2.01       0.3
4   5.4 -1.86     -18.1
5  23.5  3.14       0.0

当我进行检查以找出new_diff5.0的行时,我得到一个空的数据框.但是,当我在< 5.0> 5.0上进行检查时,它可以正常工作.见下文:

When I do a check to find out rows where new_diff is 5.0, I get an empty dataframe. But, it works fine when I do a check on < 5.0 or > 5.0. See below:

In [1567]: df[df['new_diff'] == 5.0]
Out[1567]: 
Empty DataFrame
Columns: [A, B, new_diff]
Index: []

In [1568]: df[df['new_diff'] > 5.0]
Out[1568]: 
      A     B  new_diff 
1  23.2  1.33      12.5  

In [1569]: df[df['new_diff'] < 5.0]
Out[1569]: 
      A     B  new_diff
0  16.3  1.10      -6.9
2  10.7 -0.43       5.0
3   5.7 -2.01       0.3
4   5.4 -1.86     -18.1
5  23.5  3.14       0.0

请让我知道我在这里想念什么?

Please let me know what am I missing here?

推荐答案

问题具有浮点精度,需要

Problem is with float precision, need numpy.isclose:

print (df['new_diff'].tolist())
[-6.899999999999999, 12.5, 4.999999999999999, 0.2999999999999998, -18.1, 0.0]

print (df[np.isclose(df['new_diff'], 5)])
      A     B  new_diff
2  10.7 -0.43       5.0

这篇关于 pandas :浮点数的差异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IT屋!

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