Question

我试图从列表中删除'nan'，但它拒绝去。我试过np.nan和'nan'。

这是我的代码：

ztt = []
for i in z:
    if i != 'nan':
        ztt.append(i) 
ztt

或：

ztt = []
for i in z:
    if i != np.nan:
        ztt.append(i) 
ztt

我仍然得到输出：

[[46.0, 34.0, 32.0, 40.0, 34.0, 29.0, 38.0, 39.0, 45.0, 32.0, 28.0, 43.0],
 [32.0, 30.0, 67.0, 66.0, 28.0, 19.0, 39.0, 32.0, 51.0, 28.0, 20.0, 36.0],
 [29.0, 24.0, 37.0, 31.0, 32.0, 34.0, 28.0, 31.0, 28.0, 33.0, 28.0, 39.0],
 [27.0, 29.0, 35.0, nan, nan, nan, nan, nan, nan, nan, nan, nan]]

任何人都知道出了什么问题？

Answer 1

for i in z:
    if not math.isnan(i):
        ztt.append(i)

Answer 2

nan在你的情况下DOESN＆T; T需要引号，coz nan是一个特殊的数字。当它被引用时它变成一个字符串类型。所以看起来应该是这样的：

ztt = []
for i in z:
    if !math.isnan(i)
        ztt.append(i) 
ztt

或

ztt =[value for value in z if not math.isnan(value)]

Answer 3

ztt = []
for z_i in z:
    row = []
    for z_ij in z_i:
        if math.isnan(z_ij):
            row.append(z_ij)
        # If you want to replace with, for example, 0:
        # else:
        #     row.append(0)
    ztt.append(row)

或者，使用嵌套列表推导：

ztt = [[z_ij for z_ij in zi if math.isnan(z_ij)] for z_i in z]

顺便说一下，如果您使用的是NumPy，您可能只会这样做：

import numpy as np

ztt =[value[~np.isnan(value)] for value in z]

Answer 4

您的代码的主要问题是，np.nan != np.nan是True

您也可以考虑使用数组而不是列表。

import numpy as np
z = np.array(z)   #convert into array, which allows other indexing
ztt = z[np.logical_not(np.isnan(z))]

Answer 5

为什么不在for循环中使用，即

k = [[46.0, 34.0, 32.0, 40.0, 34.0, 29.0, 38.0, 39.0, 45.0, 32.0, 28.0, 43.0],
 [32.0, 30.0, 67.0, 66.0, 28.0, 19.0, 39.0, 32.0, 51.0, 28.0, 20.0, 36.0],
 [29.0, 24.0, 37.0, 31.0, 32.0, 34.0, 28.0, 31.0, 28.0, 33.0, 28.0, 39.0],
 [27.0, 29.0, 35.0, 'nan', 'nan', 'nan', 'nan', 'nan', 'nan', 'nan', 'nan', 'nan']]

for i,j in enumerate(k):
    while 'nan' in k[i]: k[i].remove('nan')

输出：

[[46.0, 34.0, 32.0, 40.0, 34.0, 29.0, 38.0, 39.0, 45.0, 32.0, 28.0, 43.0],
 [32.0, 30.0, 67.0, 66.0, 28.0, 19.0, 39.0, 32.0, 51.0, 28.0, 20.0, 36.0],
 [29.0, 24.0, 37.0, 31.0, 32.0, 34.0, 28.0, 31.0, 28.0, 33.0, 28.0, 39.0],
 [27.0, 29.0, 35.0]]

m = [[46.0, 34.0, 32.0, 40.0, 34.0, 29.0, 38.0, 39.0, 45.0, 32.0, 28.0, 43.0],
  [32.0, 30.0, 67.0, 66.0, 28.0, 19.0, 39.0, 32.0, 51.0, 28.0, 20.0, 36.0],
  [29.0, 24.0, 37.0, 31.0, 32.0, 34.0, 28.0, 31.0, 28.0, 33.0, 28.0, 39.0],
  [27.0, 29.0, 35.0,np.nan,np.nan,np.nan,np.nan,np.nan]]

for i,j in enumerate(m):
    while np.nan in m[i]: m[i].remove(np.nan)

输出：

[[46.0, 34.0, 32.0, 40.0, 34.0, 29.0, 38.0, 39.0, 45.0, 32.0, 28.0, 43.0],
 [32.0, 30.0, 67.0, 66.0, 28.0, 19.0, 39.0, 32.0, 51.0, 28.0, 20.0, 36.0],
 [29.0, 24.0, 37.0, 31.0, 32.0, 34.0, 28.0, 31.0, 28.0, 33.0, 28.0, 39.0],
 [27.0, 29.0, 35.0]]

从列表中删除nan - Python

5 个答案: