如何在熊猫数据框架的一列中用零替换NaN值?

我有一个熊猫数据框架如下:

      itm Date                  Amount 
67    420 2012-09-30 00:00:00   65211
68    421 2012-09-09 00:00:00   29424
69    421 2012-09-16 00:00:00   29877
70    421 2012-09-23 00:00:00   30990
71    421 2012-09-30 00:00:00   61303
72    485 2012-09-09 00:00:00   71781
73    485 2012-09-16 00:00:00     NaN
74    485 2012-09-23 00:00:00   11072
75    485 2012-09-30 00:00:00  113702
76    489 2012-09-09 00:00:00   64731
77    489 2012-09-16 00:00:00     NaN

当我尝试应用一个函数到金额列，我得到以下错误:

ValueError: cannot convert float NaN to integer

我尝试使用数学模块中的.isnan应用一个函数我已经尝试了pandas .replace属性我尝试了pandas 0.9中的.sparse data属性我还尝试了在函数中if NaN == NaN语句。我也看了这篇文章我如何替换NA值与零在一个R数据框架?同时看一些其他的文章。我尝试过的所有方法都不起作用或不能识别NaN。任何提示或解决方案将不胜感激。

当前回答

我认为这也是值得提及和解释的 fillna()的参数配置如方法，轴，限制等。

从我们的文档来看:

Series.fillna(value=None, method=None, axis=None, 
                 inplace=False, limit=None, downcast=None)
Fill NA/NaN values using the specified method.

参数

value [scalar, dict, Series, or DataFrame] Value to use to 
 fill holes (e.g. 0), alternately a dict/Series/DataFrame 
 of values specifying which value to use for each index 
 (for a Series) or column (for a DataFrame). Values not in 
 the dict/Series/DataFrame will not be filled. This 
 value cannot be a list.

method [{‘backfill’, ‘bfill’, ‘pad’, ‘ffill’, None}, 
 default None] Method to use for filling holes in 
 reindexed Series pad / ffill: propagate last valid 
 observation forward to next valid backfill / bfill: 
 use next valid observation to fill gap axis 
 [{0 or ‘index’}] Axis along which to fill missing values.

inplace [bool, default False] If True, fill 
 in-place. Note: this will modify any other views
 on this object (e.g., a no-copy slice for a 
 column in a DataFrame).

limit [int,defaultNone] If method is specified, 
 this is the maximum number of consecutive NaN 
 values to forward/backward fill. In other words, 
 if there is a gap with more than this number of 
 consecutive NaNs, it will only be partially filled. 
 If method is not specified, this is the maximum 
 number of entries along the entire axis where NaNs
 will be filled. Must be greater than 0 if not None.

downcast [dict, default is None] A dict of item->dtype 
 of what to downcast if possible, or the string ‘infer’ 
 which will try to downcast to an appropriate equal 
 type (e.g. float64 to int64 if possible).

好的。让我们从方法= Parameter this开始有正向填充(ffill)和反向填充(bfill) Ffill正在复制前面的内容无缺失值。

例如:

import pandas as pd
import numpy as np
inp = [{'c1':10, 'c2':np.nan, 'c3':200}, {'c1':np.nan,'c2':110, 'c3':210}, {'c1':12,'c2':np.nan, 'c3':220},{'c1':12,'c2':130, 'c3':np.nan},{'c1':12,'c2':np.nan, 'c3':240}]
df = pd.DataFrame(inp)

  c1       c2      c3
0   10.0     NaN      200.0
1   NaN   110.0 210.0
2   12.0     NaN      220.0
3   12.0     130.0 NaN
4   12.0     NaN      240.0

填充:

df.fillna(method="ffill")

    c1     c2      c3
0   10.0      NaN 200.0
1   10.0    110.0   210.0
2   12.0    110.0   220.0
3   12.0    130.0   220.0
4   12.0    130.0   240.0

向后填:

df.fillna(method="bfill")

    c1      c2     c3
0   10.0    110.0   200.0
1   12.0    110.0   210.0
2   12.0    130.0   220.0
3   12.0    130.0   240.0
4   12.0      NaN   240.0

Axis参数帮助我们选择填充的方向:

填补方向:

填充：

Axis = 1 
Method = 'ffill'
----------->
  direction 

df.fillna(method="ffill", axis=1)

       c1   c2      c3
0   10.0     10.0   200.0
1    NaN    110.0   210.0
2   12.0     12.0   220.0
3   12.0    130.0   130.0
4   12.0    12.0    240.0

Axis = 0 # by default 
Method = 'ffill'
|
|       # direction 
|
V
e.g: # This is the ffill default
df.fillna(method="ffill", axis=0)

    c1     c2      c3
0   10.0      NaN   200.0
1   10.0    110.0   210.0
2   12.0    110.0   220.0
3   12.0    130.0   220.0
4   12.0    130.0   240.0

bfill：

axis= 0
method = 'bfill'
^
|
|
|
df.fillna(method="bfill", axis=0)

    c1     c2      c3
0   10.0    110.0   200.0
1   12.0    110.0   210.0
2   12.0    130.0   220.0
3   12.0    130.0   240.0
4   12.0      NaN   240.0

axis = 1
method = 'bfill'
<-----------
df.fillna(method="bfill", axis=1)
        c1     c2       c3
0    10.0   200.0   200.0
1   110.0   110.0   210.0
2    12.0   220.0   220.0
3    12.0   130.0     NaN
4    12.0   240.0   240.0

# alias:
#  'fill' == 'pad' 
#   bfill == backfill

极限参数:

df
    c1     c2      c3
0   10.0      NaN   200.0
1    NaN    110.0   210.0
2   12.0      NaN   220.0
3   12.0    130.0     NaN
4   12.0      NaN   240.0

只替换跨列的第一个NaN元素:

df.fillna(value = 'Unavailable', limit=1)
            c1           c2          c3
0          10.0 Unavailable       200.0
1   Unavailable       110.0       210.0
2          12.0         NaN       220.0
3          12.0       130.0 Unavailable
4          12.0         NaN       240.0

df.fillna(value = 'Unavailable', limit=2)

           c1            c2          c3
0          10.0 Unavailable       200.0
1   Unavailable       110.0       210.0
2          12.0 Unavailable       220.0
3          12.0       130.0 Unavailable
4          12.0         NaN       240.0

低垂的参数:

df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 5 entries, 0 to 4
Data columns (total 3 columns):
 #   Column  Non-Null Count  Dtype  
---  ------  --------------  -----  
 0   c1      4 non-null      float64
 1   c2      2 non-null      float64
 2   c3      4 non-null      float64
dtypes: float64(3)
memory usage: 248.0 bytes

df.fillna(method="ffill",downcast='infer').info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 5 entries, 0 to 4
Data columns (total 3 columns):
 #   Column  Non-Null Count  Dtype  
---  ------  --------------  -----  
 0   c1      5 non-null      int64  
 1   c2      4 non-null      float64
 2   c3      5 non-null      int64  
dtypes: float64(1), int64(2)
memory usage: 248.0 bytes

2022-03-29 22:38:01

其他回答

替换熊猫中的na值

df['column_name'].fillna(value_to_be_replaced,inplace=True)

如果inplace = False，它不会更新df (dataframe)，而是返回修改后的值。

2019-03-29 19:46:45

我相信DataFrame.fillna()将为您完成此工作。

链接到文档的数据框架和系列。

例子:

In [7]: df
Out[7]: 
          0         1
0       NaN       NaN
1 -0.494375  0.570994
2       NaN       NaN
3  1.876360 -0.229738
4       NaN       NaN

In [8]: df.fillna(0)
Out[8]: 
          0         1
0  0.000000  0.000000
1 -0.494375  0.570994
2  0.000000  0.000000
3  1.876360 -0.229738
4  0.000000  0.000000

若要仅在一列中填充nan，请仅选择该列。在这种情况下，我使用inplace=True来实际改变df的内容。

In [12]: df[1].fillna(0, inplace=True)
Out[12]: 
0    0.000000
1    0.570994
2    0.000000
3   -0.229738
4    0.000000
Name: 1

In [13]: df
Out[13]: 
          0         1
0       NaN  0.000000
1 -0.494375  0.570994
2       NaN  0.000000
3  1.876360 -0.229738
4       NaN  0.000000

编辑:

为了避免SettingWithCopyWarning，使用内置的列特定功能:

df.fillna({1:0}, inplace=True)

2012-11-08 18:54:27

主要有两种选择;在输入或填充缺失值NaN / np时。只有数值替换的Nan(跨列):

df(“金额”)。fillna(value=None, method=，axis=1，)是足够的:

来自文档:

取值:scalar、dict、Series或DataFrame 值用于填充孔(例如0)，交替使用a dict/Series/DataFrame的值，指定用于哪个值每个索引(对于Series)或列(对于DataFrame)。(值不在dict/Series/DataFrame中将不会被填充)。此值不能列个清单。

这意味着'字符串'或'常量'不再允许被赋值。

对于更专门的imputer，使用SimpleImputer():

from sklearn.impute import SimpleImputer
si = SimpleImputer(strategy='constant', missing_values=np.nan, fill_value='Replacement_Value')
df[['Col-1', 'Col-2']] = si.fit_transform(X=df[['C-1', 'C-2']])

2019-07-23 13:36:14

并不保证切片返回一个视图或副本。你可以这样做

df['column'] = df['column'].fillna(value)

2016-10-06 09:10:08