pandas DataFrame:用列的平均值替换nan值

问题:pandas DataFrame:用列的平均值替换nan值

我有一个熊猫DataFrame,其中大多数都是实数,但其中也有一些nan值。

如何nan用列的平均值替换s?

这个问题与这个问题非常相似:numpy array:用列的平均值替换nan值, 但是不幸的是,给出的解决方案不适用于pandas DataFrame。

I’ve got a pandas DataFrame filled mostly with real numbers, but there is a few nan values in it as well.

How can I replace the nans with averages of columns where they are?

This question is very similar to this one: numpy array: replace nan values with average of columns but, unfortunately, the solution given there doesn’t work for a pandas DataFrame.


回答 0

您可以直接使用DataFrame.fillnanan直接填充:

In [27]: df 
Out[27]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3       NaN -2.027325  1.533582
4       NaN       NaN  0.461821
5 -0.788073       NaN       NaN
6 -0.916080 -0.612343       NaN
7 -0.887858  1.033826       NaN
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

In [28]: df.mean()
Out[28]: 
A   -0.151121
B   -0.231291
C   -0.530307
dtype: float64

In [29]: df.fillna(df.mean())
Out[29]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 -0.151121 -2.027325  1.533582
4 -0.151121 -0.231291  0.461821
5 -0.788073 -0.231291 -0.530307
6 -0.916080 -0.612343 -0.530307
7 -0.887858  1.033826 -0.530307
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

的文档字符串fillna说,value应该是一个标量或快译通,但是,它似乎工作用Series为好。如果您想通过字典,可以使用df.mean().to_dict()

You can simply use DataFrame.fillna to fill the nan‘s directly:

In [27]: df 
Out[27]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3       NaN -2.027325  1.533582
4       NaN       NaN  0.461821
5 -0.788073       NaN       NaN
6 -0.916080 -0.612343       NaN
7 -0.887858  1.033826       NaN
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

In [28]: df.mean()
Out[28]: 
A   -0.151121
B   -0.231291
C   -0.530307
dtype: float64

In [29]: df.fillna(df.mean())
Out[29]: 
          A         B         C
0 -0.166919  0.979728 -0.632955
1 -0.297953 -0.912674 -1.365463
2 -0.120211 -0.540679 -0.680481
3 -0.151121 -2.027325  1.533582
4 -0.151121 -0.231291  0.461821
5 -0.788073 -0.231291 -0.530307
6 -0.916080 -0.612343 -0.530307
7 -0.887858  1.033826 -0.530307
8  1.948430  1.025011 -2.982224
9  0.019698 -0.795876 -0.046431

The docstring of fillna says that value should be a scalar or a dict, however, it seems to work with a Series as well. If you want to pass a dict, you could use df.mean().to_dict().


回答 1

尝试:

sub2['income'].fillna((sub2['income'].mean()), inplace=True)

Try:

sub2['income'].fillna((sub2['income'].mean()), inplace=True)

回答 2

In [16]: df = DataFrame(np.random.randn(10,3))

In [17]: df.iloc[3:5,0] = np.nan

In [18]: df.iloc[4:6,1] = np.nan

In [19]: df.iloc[5:8,2] = np.nan

In [20]: df
Out[20]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3       NaN -0.985188 -0.324136
4       NaN       NaN  0.238512
5  0.769657       NaN       NaN
6  0.141951  0.326064       NaN
7 -1.694475 -0.523440       NaN
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794

In [22]: df.mean()
Out[22]: 
0   -0.251534
1   -0.040622
2   -0.841219
dtype: float64

应用每列该列的平均值并填充

In [23]: df.apply(lambda x: x.fillna(x.mean()),axis=0)
Out[23]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3 -0.251534 -0.985188 -0.324136
4 -0.251534 -0.040622  0.238512
5  0.769657 -0.040622 -0.841219
6  0.141951  0.326064 -0.841219
7 -1.694475 -0.523440 -0.841219
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794
In [16]: df = DataFrame(np.random.randn(10,3))

In [17]: df.iloc[3:5,0] = np.nan

In [18]: df.iloc[4:6,1] = np.nan

In [19]: df.iloc[5:8,2] = np.nan

In [20]: df
Out[20]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3       NaN -0.985188 -0.324136
4       NaN       NaN  0.238512
5  0.769657       NaN       NaN
6  0.141951  0.326064       NaN
7 -1.694475 -0.523440       NaN
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794

In [22]: df.mean()
Out[22]: 
0   -0.251534
1   -0.040622
2   -0.841219
dtype: float64

Apply per-column the mean of that columns and fill

In [23]: df.apply(lambda x: x.fillna(x.mean()),axis=0)
Out[23]: 
          0         1         2
0  1.148272  0.227366 -2.368136
1 -0.820823  1.071471 -0.784713
2  0.157913  0.602857  0.665034
3 -0.251534 -0.985188 -0.324136
4 -0.251534 -0.040622  0.238512
5  0.769657 -0.040622 -0.841219
6  0.141951  0.326064 -0.841219
7 -1.694475 -0.523440 -0.841219
8  0.352556 -0.551487 -1.639298
9 -2.067324 -0.492617 -1.675794

回答 3

# To read data from csv file
Dataset = pd.read_csv('Data.csv')

X = Dataset.iloc[:, :-1].values

# To calculate mean use imputer class
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(missing_values=np.nan, strategy='mean')
imputer = imputer.fit(X[:, 1:3])
X[:, 1:3] = imputer.transform(X[:, 1:3])
# To read data from csv file
Dataset = pd.read_csv('Data.csv')

X = Dataset.iloc[:, :-1].values

# To calculate mean use imputer class
from sklearn.impute import SimpleImputer
imputer = SimpleImputer(missing_values=np.nan, strategy='mean')
imputer = imputer.fit(X[:, 1:3])
X[:, 1:3] = imputer.transform(X[:, 1:3])

回答 4

如果您想用均值来估算缺失值,并且想逐列进行计算,则只会用该列的均值来估算。这可能更具可读性。

sub2['income'] = sub2['income'].fillna((sub2['income'].mean()))

If you want to impute missing values with mean and you want to go column by column, then this will only impute with the mean of that column. This might be a little more readable.

sub2['income'] = sub2['income'].fillna((sub2['income'].mean()))

回答 5

直接使用df.fillna(df.mean())均值填充所有空值

如果要用该列的平均值填充空值,则可以使用此值

假设x=df['Item_Weight']这里Item_Weight是列名

这是我们要分配的(将x的空值和x的平均值填充到x中)

df['Item_Weight'] = df['Item_Weight'].fillna((df['Item_Weight'].mean()))

如果要用某些字符串填充空值,请使用

Outlet_size是列名

df.Outlet_Size = df.Outlet_Size.fillna('Missing')

Directly use df.fillna(df.mean()) to fill all the null value with mean

If you want to fill null value with mean of that column then you can use this

suppose x=df['Item_Weight'] here Item_Weight is column name

here we are assigning (fill null values of x with mean of x into x)

df['Item_Weight'] = df['Item_Weight'].fillna((df['Item_Weight'].mean()))

If you want to fill null value with some string then use

here Outlet_size is column name

df.Outlet_Size = df.Outlet_Size.fillna('Missing')

回答 6

除上述之外,另一个选择是:

df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))

它的平均值不如以前的平均值那么优雅,但是如果您希望用其他某些列函数替换空值,它可能会更短。

Another option besides those above is:

df = df.groupby(df.columns, axis = 1).transform(lambda x: x.fillna(x.mean()))

It’s less elegant than previous responses for mean, but it could be shorter if you desire to replace nulls by some other column function.


回答 7

熊猫:如何用nan一栏的平均值(均值),中位数或其他统计量替换NaN()值

假设您的DataFrame是,df并且您有一列称为nr_items。这是: df['nr_items']

如果要用列的平均值替换NaN列的值:df['nr_items']

使用方法.fillna()

mean_value=df['nr_items'].mean()
df['nr_item_ave']=df['nr_items'].fillna(mean_value)

我创建了一个新df列,称为nr_item_ave存储新列,其中的NaN值替换mean为该列的值。

使用时应小心mean。如果您有异常值,建议使用median

Pandas: How to replace NaN (nan) values with the average (mean), median or other statistics of one column

Say your DataFrame is df and you have one column called nr_items. This is: df['nr_items']

If you want to replace the NaN values of your column df['nr_items'] with the mean of the column:

Use method .fillna():

mean_value=df['nr_items'].mean()
df['nr_item_ave']=df['nr_items'].fillna(mean_value)

I have created a new df column called nr_item_ave to store the new column with the NaN values replaced by the mean value of the column.

You should be careful when using the mean. If you have outliers is more recommendable to use the median


回答 8

使用sklearn库预处理类

from sklearn.impute import SimpleImputer
missingvalues = SimpleImputer(missing_values = np.nan, strategy = 'mean', axis = 0)
missingvalues = missingvalues.fit(x[:,1:3])
x[:,1:3] = missingvalues.transform(x[:,1:3])

注意:在最新版本中,参数missing_values值更改为np.nanfromNaN

using sklearn library preprocessing class

from sklearn.impute import SimpleImputer
missingvalues = SimpleImputer(missing_values = np.nan, strategy = 'mean', axis = 0)
missingvalues = missingvalues.fit(x[:,1:3])
x[:,1:3] = missingvalues.transform(x[:,1:3])

Note: In the recent version parameter missing_values value change to np.nan from NaN