Question

所以我研究了多种可能的解决方案，但似乎都没有效果。

基本上，我想在我的数据框中创建一个新列，它是多个其他列的平均值。我希望这个均值排除 NaN 值，但即使行中有 NaN 值，仍然计算均值。

我有一个看起来像这样的数据框（但实际上是 Q222-229）：

ID   Q1   Q2   Q3   Q4   Q5
1    4    NaN  NaN  NaN  NaN
2    5    7    8    NaN  NaN
3    7    1    2    NaN  NaN
4    2    2    3    4    1
5    1    3    NaN  NaN  NaN

我想创建一个列，它是 Q1、Q2、Q3、Q4、Q5 的平均值，即：

ID   Q1   Q2   Q3   Q4   Q5   avg_age
1    4    NaN  NaN  NaN  NaN  4
2    5    7    8    NaN  NaN  5.5
3    7    1    2    NaN  NaN  3.5
4    2    2    3    4    1    2
5    1    3    NaN  NaN  NaN  2

（忽略值）

但是，我尝试过的每种方法都会在 avg_age 列中返回 NaN 值，这让我认为在忽略 NaN 值时，pandas 会忽略整行。但我不希望这种情况发生，而是希望在忽略 NaN 值的情况下返回均值。

这是我迄今为止尝试过的：

1.
    avg_age = s.loc[: , "Q222":"Q229"]
    avg_age = avg_age.mean(axis=1)
    s = pd.concat([s, avg_age], axis=1)

2.
    s['avg_age'] = s[['Q222', 'Q223', 'Q224', 'Q225', 'Q226', 'Q227', 'Q228', 'Q229']].mean(axis=1)

3.

    avg_age = ['Q222', 'Q223', 'Q224', 'Q225', 'Q226', 'Q227', 'Q228', 'Q229']
    s.loc[:, 'avg_age'] = s[avg_age].mean(axis=1)

我不确定我最初对值进行编码的方式是否有问题，所以这是我的代码供参考：

#Changing age 变量输入

s['Q222'] = s['Q222'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q223'] = s['Q223'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q224'] = s['Q224'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q225'] = s['Q225'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q226'] = s['Q226'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q227'] = s['Q227'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q228'] = s['Q228'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])
s['Q229'] = s['Q229'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                              ['2','3','4','5', '6', '7', '8', np.NaN])

s['Q222'] = s['Q222'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q223'] = s['Q223'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q224'] = s['Q224'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q225'] = s['Q225'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q226'] = s['Q226'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q227'] = s['Q227'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q228'] = s['Q228'].replace(['0-4', '05-11', '12-15', '16-17'], '1')
s['Q229'] = s['Q229'].replace(['0-4', '05-11', '12-15', '16-17'], '1')

提前感谢任何能够提供帮助的人！

Answer 1

skipna=True

可以使用 list comprehension 获取列的平均值，并使用 mean() 获取：

df['ave_age'] = df[[col for col in df.columns if 'Q' in col]].mean(axis = 1,skipna = True)

Answer 2

DataFrame.mean() 的默认行为应该按照您的意愿行事。

以下示例显示对列的子集取平均值并将其放入新创建的列中：

In[19]: tmp
Out[19]: 
   a  b    c
0  1  2  5.0
1  2  3  6.0
2  3  4  NaN

In[24]: tmp['mean'] = tmp[['b', 'c']].mean(axis=1)

In[25]: tmp
Out[25]: 
   a  b    c  mean
0  1  2  5.0   3.5
1  2  3  6.0   4.5
2  3  4  NaN   4.0

至于你的代码出了什么问题：

<块引用>

s['Q222'] = s['Q222'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                         ['2','3','4','5', '6', '7', '8', np.NaN])

您的数据框中没有数值（即 2、3、4），您有字符串（'2'、'3' 和 '4'）。 DataFrame.mean() 函数将这些字符串视为 NaN，因此您将获得 NaN 作为所有均值计算的结果。

尝试用数字填充您的框架，如下所示：

 s['Q222'] = s['Q222'].replace(['18-24', '25-34','35-44', '45-54','55-64', '65-74', '75 or older', "Don't know"],
                          [2, 3, 4, 5, 6, 7, 8, np.NaN])

创建一个列，它是 Pandas 中数据框中多列的平均值

2 个答案: