Python 如何检查包含 NaN 的列表
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How to Check list containing NaN
提问by user1234440
In my for loop, my code generates a list like this one:
在我的 for 循环中，我的代码生成了一个这样的列表：
list([0.0,0.0]/sum([0.0,0.0]))
The loop generates all sort of other number vectors but it also generates [nan,nan], and to avoid it I tried to put in a conditional to prevent it like the one below, but it doesn't return true.
该循环生成各种其他数字向量，但它也会生成[nan,nan]，为了避免它，我尝试放入一个条件来防止它，如下所示，但它没有返回 true。
nan in list([0.0,0.0]/sum([0.0,0.0]))
>>> False
Shouldn't it return true?
它不应该返回true吗？
Libraries I've loaded:
我加载的库：
import PerformanceAnalytics as perf
import DataAnalyticsHelpers
import DataHelpers as data
import OptimizationHelpers as optim
from matplotlib.pylab import *
from pandas.io.data import DataReader
from datetime import datetime,date,time
import tradingWithPython as twp
import tradingWithPython.lib.yahooFinance as data_downloader # used to get data from yahoo finance
import pandas as pd # as always.
import numpy as np
import zipline as zp
from scipy.optimize import minimize
from itertools import product, combinations
import time
from math import isnan
采纳答案by DSM
I think this makes sense because of your pulling numpyinto scope indirectly via the star import.
我认为这是有道理的，因为您numpy通过星形导入间接进入了范围。
>>> import numpy as np
>>> [0.0,0.0]/0
Traceback (most recent call last):
File "<ipython-input-3-aae9e30b3430>", line 1, in <module>
[0.0,0.0]/0
TypeError: unsupported operand type(s) for /: 'list' and 'int'
>>> [0.0,0.0]/np.float64(0)
array([ nan, nan])
When you did
当你做
from matplotlib.pylab import *
it pulled in numpy.sum:
它拉进来numpy.sum：
>>> from matplotlib.pylab import *
>>> sum is np.sum
True
>>> [0.0,0.0]/sum([0.0, 0.0])
array([ nan, nan])
You can test that thisnanobject (nanisn't unique in general) is in a list via identity, but if you try it in an arrayit seems to test via equality, and nan != nan:
您可以通过身份测试此nan对象（nan通常不是唯一的）是否在列表中，但是如果您在一个列表中尝试array它似乎通过相等性进行测试，并且nan != nan：
>>> nan == nan
False
>>> nan == nan, nan is nan
(False, True)
>>> nan in [nan]
True
>>> nan in np.array([nan])
False
You could use np.isnan:
你可以使用np.isnan：
>>> np.isnan([nan, nan])
array([ True, True], dtype=bool)
>>> np.isnan([nan, nan]).any()
True
回答by AbhishekLohade
May be this is what you are looking for...
可能这就是你正在寻找的......
a = [2,3,np.nan]
b = True if True in np.isnan(np.array(a)) else False
print(b)
