pandas 忽略nan的Python比较
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Python comparison ignoring nan
提问by sds
While nan == nanis always False, in many cases people want to treat them as equal, and this is enshrined in pandas.DataFrame.equals:
虽然nan == nan总是False，但在许多情况下，人们希望将它们视为平等，这体现在pandas.DataFrame.equals：
NaNs in the same location are considered equal.
同一位置的 NaN 被认为是相等的。
Of course, I can write
当然，我可以写
def equalp(x, y):
return (x == y) or (math.isnan(x) and math.isnan(y))
However, this will fail on containers like [float("nan")]and isnanbarfs on non-numbers (so the complexity increases).
但是，这将在非数字上的容器[float("nan")]和isnanbarfs 上失败（因此复杂性增加）。
So, what do people do to compare complex Python objects which may contain nan?
那么，人们如何比较可能包含的复杂 Python 对象nan呢？
PS. Motivation: when comparing two rows in a pandas DataFrame, I would convert them into dictsand compare dicts element-wise.
附注。动机：当比较 pandas 中的两行时DataFrame，我会将它们转换为dicts并按元素比较 dicts 。
采纳答案by juanpa.arrivillaga
Suppose you have a data-frame with nanvalues:
假设您有一个包含nan值的数据框：
In [10]: df = pd.DataFrame(np.random.randint(0, 20, (10, 10)).astype(float), columns=["c%d"%d for d in range(10)])
In [10]: df.where(np.random.randint(0,2, df.shape).astype(bool), np.nan, inplace=True)
In [10]: df
Out[10]:
c0 c1 c2 c3 c4 c5 c6 c7 c8 c9
0 NaN 6.0 14.0 NaN 5.0 NaN 2.0 12.0 3.0 7.0
1 NaN 6.0 5.0 17.0 NaN NaN 13.0 NaN NaN NaN
2 NaN 17.0 NaN 8.0 6.0 NaN NaN 13.0 NaN NaN
3 3.0 NaN NaN 15.0 NaN 8.0 3.0 NaN 3.0 NaN
4 7.0 8.0 7.0 NaN 9.0 19.0 NaN 0.0 NaN 11.0
5 NaN NaN 14.0 2.0 NaN NaN 0.0 NaN NaN 8.0
6 3.0 13.0 NaN NaN NaN NaN NaN 12.0 3.0 NaN
7 13.0 14.0 NaN 5.0 13.0 NaN 18.0 6.0 NaN 5.0
8 3.0 9.0 14.0 19.0 11.0 NaN NaN NaN NaN 5.0
9 3.0 17.0 NaN NaN 0.0 NaN 11.0 NaN NaN 0.0
And you want to compare rows, say, row 0 and 8. Then just use fillnaand do vectorized comparison:
并且您想比较行，例如第 0 行和第 8 行。然后只需使用fillna并进行矢量化比较：
In [12]: df.iloc[0,:].fillna(0) != df.iloc[8,:].fillna(0)
Out[12]:
c0 True
c1 True
c2 False
c3 True
c4 True
c5 False
c6 True
c7 True
c8 True
c9 True
dtype: bool
You can use the resulting boolean array to index into the columns, if you just want to know which columns are different:
如果您只想知道哪些列不同，您可以使用生成的布尔数组对列进行索引：
In [14]: df.columns[df.iloc[0,:].fillna(0) != df.iloc[8,:].fillna(0)]
Out[14]: Index(['c0', 'c1', 'c3', 'c4', 'c6', 'c7', 'c8', 'c9'], dtype='object')
回答by ascripter
I assume you have array-data or can at least convert to a numpy array?
我假设您有数组数据或至少可以转换为 numpy 数组？
One way is to mask all the nans using a numpy.maarray, then comparing the arrays. So your starting situation would be sth. like this
一种方法是使用numpy.ma数组屏蔽所有 nan ，然后比较数组。所以你的开始情况会是…… 像这样
import numpy as np
import numpy.ma as ma
arr1 = ma.array([3,4,6,np.nan,2])
arr2 = ma.array([3,4,6,np.nan,2])
print arr1 == arr2
print ma.all(arr1==arr2)
>>> [ True True True False True]
>>> False # <-- you want this to show True
Solution:
解决方案：
arr1[np.isnan(arr1)] = ma.masked
arr2[np.isnan(arr2)] = ma.masked
print arr1 == arr2
print ma.all(arr1==arr2)
>>> [True True True -- True]
>>> True
