Python 忽略 nan 的比较

Question

虽然 nan == nan 始终是 False，但在很多情况下人们希望平等对待他们，这体现在 pandas.DataFrame.equals:

NaNs in the same location are considered equal.

当然可以写

def equalp(x, y):
    return (x == y) or (math.isnan(x) and math.isnan(y))

但是，这将在 [float("nan")] 和 isnan 等非数字 barfs 容器上失败（因此 the complexity increases）。

那么，人们如何比较可能包含 nan 的复杂 Python 对象？

PS。动机：当比较 pandas DataFrame 中的两行时，我会 convert them into dicts 并按元素比较字典。

PPS。当我说“compare”时，我在想 diff, not equalp.

Answer 1

我假设你有数组数据或者至少可以转换为 numpy 数组？

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

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

Answer 2

假设您有一个包含 nan 个值的数据框：

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

你想比较行，比如第 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

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')

Answer 3

这是一个递归到数据结构中的函数，用唯一字符串替换 nan 值。我写这篇文章是为了比较可能包含 nan.

的数据结构的单元测试

它只是为dict和list构成的数据结构而设计的，但是很容易看出如何扩展它。

from math import isnan
from uuid import uuid4
from typing import Union

NAN_REPLACEMENT = f"THIS_WAS_A_NAN{uuid4()}"

def replace_nans(data_structure: Union[dict, list]) -> Union[dict, list]:
    if isinstance(data_structure, dict):
        iterme = data_structure.items()
    elif isinstance(data_structure, list):
        iterme = enumerate(data_structure)
    else:
        raise ValueError(
            "replace_nans should only be called on structures made of dicts and lists"
        )

    for key, value in iterme:
        if isinstance(value, float) and isnan(value):
            data_structure[key] = NAN_REPLACEMENT
        elif isinstance(value, dict) or isinstance(value, list):
            data_structure[key] = replace_nans(data_structure[key])
    return data_structure

Python 忽略 nan 的比较

Python comparison ignoring nan

python

equality

nan

python-2.7

pandas