如何将熊猫数据框的索引转换为列?

问题:如何将熊猫数据框的索引转换为列?

这似乎很明显,但是我似乎无法弄清楚如何将数据帧的索引转换为列?

例如:

df=
        gi       ptt_loc
 0  384444683      593  
 1  384444684      594 
 2  384444686      596  

至,

df=
    index1    gi       ptt_loc
 0  0     384444683      593  
 1  1     384444684      594 
 2  2     384444686      596  

This seems rather obvious, but I can’t seem to figure out how to convert an index of data frame to a column?

For example:

df=
        gi       ptt_loc
 0  384444683      593  
 1  384444684      594 
 2  384444686      596  

To,

df=
    index1    gi       ptt_loc
 0  0     384444683      593  
 1  1     384444684      594 
 2  2     384444686      596  

回答 0

要么:

df['index1'] = df.index

.reset_index

df.reset_index(level=0, inplace=True)

因此,如果您有一个3级索引的多索引框架,例如:

>>> df
                       val
tick       tag obs        
2016-02-26 C   2    0.0139
2016-02-27 A   2    0.5577
2016-02-28 C   6    0.0303

并且要将索引中的第1级(tick)和第3级(obs)转换为列,您可以执行以下操作:

>>> df.reset_index(level=['tick', 'obs'])
          tick  obs     val
tag                        
C   2016-02-26    2  0.0139
A   2016-02-27    2  0.5577
C   2016-02-28    6  0.0303

either:

df['index1'] = df.index

or, .reset_index:

df.reset_index(level=0, inplace=True)

so, if you have a multi-index frame with 3 levels of index, like:

>>> df
                       val
tick       tag obs        
2016-02-26 C   2    0.0139
2016-02-27 A   2    0.5577
2016-02-28 C   6    0.0303

and you want to convert the 1st (tick) and 3rd (obs) levels in the index into columns, you would do:

>>> df.reset_index(level=['tick', 'obs'])
          tick  obs     val
tag                        
C   2016-02-26    2  0.0139
A   2016-02-27    2  0.5577
C   2016-02-28    6  0.0303

回答 1

对于MultiIndex,您可以使用以下方法提取其子索引

df['si_name'] = R.index.get_level_values('si_name') 

si_name索引的名称在哪里。

For MultiIndex you can extract its subindex using

df['si_name'] = R.index.get_level_values('si_name') 

where si_name is the name of the subindex.


回答 2

为了更加清楚,让我们看一下索引中具有两个级别的DataFrame(一个MultiIndex)。

index = pd.MultiIndex.from_product([['TX', 'FL', 'CA'], 
                                    ['North', 'South']], 
                                   names=['State', 'Direction'])

df = pd.DataFrame(index=index, 
                  data=np.random.randint(0, 10, (6,4)), 
                  columns=list('abcd'))

reset_index使用默认参数调用的方法将所有索引级别转换为列,并使用简单RangeIndex的新索引。

df.reset_index()

使用level参数控制将哪些索引级别转换为列。如果可能,请使用更明确的级别名称。如果没有级别名称,则可以通过其整数位置来引用每个级别,整数位置从外部开始为0。您可以在此处使用标量值或要重置的所有索引的列表。

df.reset_index(level='State') # same as df.reset_index(level=0)

在极少数情况下,您想要保留索引并将索引转换为列,可以执行以下操作:

# for a single level
df.assign(State=df.index.get_level_values('State'))

# for all levels
df.assign(**df.index.to_frame())

To provide a bit more clarity, let’s look at a DataFrame with two levels in its index (a MultiIndex).

index = pd.MultiIndex.from_product([['TX', 'FL', 'CA'], 
                                    ['North', 'South']], 
                                   names=['State', 'Direction'])

df = pd.DataFrame(index=index, 
                  data=np.random.randint(0, 10, (6,4)), 
                  columns=list('abcd'))

The reset_index method, called with the default parameters, converts all index levels to columns and uses a simple RangeIndex as new index.

df.reset_index()

Use the level parameter to control which index levels are converted into columns. If possible, use the level name, which is more explicit. If there are no level names, you can refer to each level by its integer location, which begin at 0 from the outside. You can use a scalar value here or a list of all the indexes you would like to reset.

df.reset_index(level='State') # same as df.reset_index(level=0)

In the rare event that you want to preserve the index and turn the index into a column, you can do the following:

# for a single level
df.assign(State=df.index.get_level_values('State'))

# for all levels
df.assign(**df.index.to_frame())

回答 3

rename_axis + reset_index

您可以先将索引重命名为所需的标签,然后提升为一系列:

df = df.rename_axis('index1').reset_index()

print(df)

   index1         gi  ptt_loc
0       0  384444683      593
1       1  384444684      594
2       2  384444686      596

这也适用于MultiIndex数据框:

print(df)
#                        val
# tick       tag obs        
# 2016-02-26 C   2    0.0139
# 2016-02-27 A   2    0.5577
# 2016-02-28 C   6    0.0303

df = df.rename_axis(['index1', 'index2', 'index3']).reset_index()

print(df)

       index1 index2  index3     val
0  2016-02-26      C       2  0.0139
1  2016-02-27      A       2  0.5577
2  2016-02-28      C       6  0.0303

rename_axis + reset_index

You can first rename your index to a desired label, then elevate to a series:

df = df.rename_axis('index1').reset_index()

print(df)

   index1         gi  ptt_loc
0       0  384444683      593
1       1  384444684      594
2       2  384444686      596

This works also for MultiIndex dataframes:

print(df)
#                        val
# tick       tag obs        
# 2016-02-26 C   2    0.0139
# 2016-02-27 A   2    0.5577
# 2016-02-28 C   6    0.0303

df = df.rename_axis(['index1', 'index2', 'index3']).reset_index()

print(df)

       index1 index2  index3     val
0  2016-02-26      C       2  0.0139
1  2016-02-27      A       2  0.5577
2  2016-02-28      C       6  0.0303

回答 4

如果要使用该reset_index方法并保留现有索引,则应使用:

df.reset_index().set_index('index', drop=False)

或更改它的位置:

df.reset_index(inplace=True)
df.set_index('index', drop=False, inplace=True)

例如:

print(df)
          gi  ptt_loc
0  384444683      593
4  384444684      594
9  384444686      596

print(df.reset_index())
   index         gi  ptt_loc
0      0  384444683      593
1      4  384444684      594
2      9  384444686      596

print(df.reset_index().set_index('index', drop=False))
       index         gi  ptt_loc
index
0          0  384444683      593
4          4  384444684      594
9          9  384444686      596

如果要摆脱索引标签,可以执行以下操作:

df2 = df.reset_index().set_index('index', drop=False)
df2.index.name = None
print(df2)
   index         gi  ptt_loc
0      0  384444683      593
4      4  384444684      594
9      9  384444686      596

If you want to use the reset_index method and also preserve your existing index you should use:

df.reset_index().set_index('index', drop=False)

or to change it in place:

df.reset_index(inplace=True)
df.set_index('index', drop=False, inplace=True)

For example:

print(df)
          gi  ptt_loc
0  384444683      593
4  384444684      594
9  384444686      596

print(df.reset_index())
   index         gi  ptt_loc
0      0  384444683      593
1      4  384444684      594
2      9  384444686      596

print(df.reset_index().set_index('index', drop=False))
       index         gi  ptt_loc
index
0          0  384444683      593
4          4  384444684      594
9          9  384444686      596

And if you want to get rid of the index label you can do:

df2 = df.reset_index().set_index('index', drop=False)
df2.index.name = None
print(df2)
   index         gi  ptt_loc
0      0  384444683      593
4      4  384444684      594
9      9  384444686      596

回答 5

df1 = pd.DataFrame({"gi":[232,66,34,43],"ptt":[342,56,662,123]})
p = df1.index.values
df1.insert( 0, column="new",value = p)
df1

    new     gi     ptt
0    0      232    342
1    1      66     56 
2    2      34     662
3    3      43     123
df1 = pd.DataFrame({"gi":[232,66,34,43],"ptt":[342,56,662,123]})
p = df1.index.values
df1.insert( 0, column="new",value = p)
df1

    new     gi     ptt
0    0      232    342
1    1      66     56 
2    2      34     662
3    3      43     123

回答 6

一种简单的方法是使用reset_index()方法。对于数据帧df,请使用以下代码:

df.reset_index(inplace=True)

这样,索引将成为一列,并且通过使用inplace作为True,这将成为永久更改。

A very simple way of doing this is to use reset_index() method.For a data frame df use the code below:

df.reset_index(inplace=True)

This way, the index will become a column, and by using inplace as True,this become permanent change.