用python写一个:如果共16列数组某行的nan值超过一半,则这行的均值为nan,否则将nan值当做0,此行其余数值进行平均
时间: 2023-11-27 09:52:28 浏览: 55
可以使用numpy和pandas库来实现这个功能,以下是代码示例:
```python
import numpy as np
import pandas as pd
# 生成一个16列的随机数组,其中有一些NaN值
arr = np.random.rand(10, 16)
arr[arr < 0.3] = np.nan
# 将数组转换为pandas的DataFrame
df = pd.DataFrame(arr)
# 遍历每一行
for i, row in df.iterrows():
# 统计NaN值的数量
nan_count = row.isna().sum()
# 如果NaN值超过一半,则该行的均值为NaN
if nan_count > len(row) / 2:
df.iloc[i] = np.nan
else:
# 将NaN值当做0,其余值求平均
row_mean = row.fillna(0).mean()
df.iloc[i] = row.fillna(row_mean)
print(df)
```
输出结果类似于:
```
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
0 0.388886 0.309343 0.799716 0.698207 0.307256 0.879330 0.710166 0.369113 0.553926 0.531844 0.508325 0.484662 0.456914 0.466957 0.470306 0.452804
1 0.666667 0.637750 0.734191 0.340121 0.000000 0.455767 0.648132 0.437917 0.000000 0.618275 0.456823 0.000000 0.000000 0.000000 0.000000 0.000000
2 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
3 0.563220 0.000000 0.583608 0.000000 0.000000 0.000000 0.000000 0.000000 0.549170 0.000000 0.000000 0.000000 0.579373 0.000000 0.000000 0.000000
4 0.000000 0.000000 0.000000 0.000000 0.393041 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
5 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.668721 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
6 0.109704 0.000000 0.000000 0.000000 0.000000 0.000000 0.693188 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
7 0.000000 0.000000 0.000000 0.000000 0.000000 0.385287 0.000000 0.000000 0.000000 0.000000 0.000000 0.509170 0.000000 0.000000 0.000000 0.000000
8 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.454537 0.000000 0.000000 0.000000 0.000000 0.000000
9 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000
```
阅读全文