缺失数据，在 Pandas 中插入行并用 NAN 填充

Question

I'm new to Python and Pandas so there might be a simple solution which I don't see.

I have a number of discontinuous datasets which look like this:

ind A    B  C  
0   0.0  1  3  
1   0.5  4  2  
2   1.0  6  1  
3   3.5  2  0  
4   4.0  4  5  
5   4.5  3  3  

I now look for a solution to get the following:

ind A    B  C  
0   0.0  1  3  
1   0.5  4  2  
2   1.0  6  1  
3   1.5  NAN NAN  
4   2.0  NAN NAN  
5   2.5  NAN NAN  
6   3.0  NAN NAN  
7   3.5  2  0  
8   4.0  4  5  
9   4.5  3  3  

The problem is,that the gap in A varies from dataset to dataset in position and length...

Answer 1

set_index and reset_index are your friends.

df = DataFrame({"A":[0,0.5,1.0,3.5,4.0,4.5], "B":[1,4,6,2,4,3], "C":[3,2,1,0,5,3]})

First move column A to the index:

In [64]: df.set_index("A")
Out[64]: 
     B  C
 A        
0.0  1  3
0.5  4  2
1.0  6  1
3.5  2  0
4.0  4  5
4.5  3  3

Then reindex with a new index, here the missing data is filled in with nans. We use the Index object since we can name it; this will be used in the next step.

In [66]: new_index = Index(arange(0,5,0.5), name="A")
In [67]: df.set_index("A").reindex(new_index)
Out[67]: 
      B   C
0.0   1   3
0.5   4   2
1.0   6   1
1.5 NaN NaN
2.0 NaN NaN
2.5 NaN NaN
3.0 NaN NaN
3.5   2   0
4.0   4   5
4.5   3   3

Finally move the index back to the columns with reset_index . Since we named the index, it all works magically:

In [69]: df.set_index("A").reindex(new_index).reset_index()
Out[69]: 
       A   B   C
0    0.0   1   3
1    0.5   4   2
2    1.0   6   1
3    1.5 NaN NaN
4    2.0 NaN NaN
5    2.5 NaN NaN
6    3.0 NaN NaN
7    3.5   2   0
8    4.0   4   5
9    4.5   3   3

Answer 2

Using the answer by EdChum above, I created the following function

def fill_missing_range(df, field, range_from, range_to, range_step=1, fill_with=0):
    return df\
      .merge(how='right', on=field,
            right = pd.DataFrame({field:np.arange(range_from, range_to, range_step)}))\
      .sort_values(by=field).reset_index().fillna(fill_with).drop(['index'], axis=1)

Example usage:

fill_missing_range(df, 'A', 0.0, 4.5, 0.5, np.nan)

Answer 3

In this case I am overwriting your A column with a newly generated dataframe and merging this to your original df, I then resort it:

    In [177]:

df.merge(how='right', on='A', right = pd.DataFrame({'A':np.arange(df.iloc[0]['A'], df.iloc[-1]['A'] + 0.5, 0.5)})).sort(columns='A').reset_index().drop(['index'], axis=1)
Out[177]:
     A   B   C
0  0.0   1   3
1  0.5   4   2
2  1.0   6   1
3  1.5 NaN NaN
4  2.0 NaN NaN
5  2.5 NaN NaN
6  3.0 NaN NaN
7  3.5   2   0
8  4.0   4   5
9  4.5   3   3

So in the general case you can adjust the arange function which takes a start and end value, note I added 0.5 to the end as ranges are open closed, and pass a step value.

A more general method could be like this:

In [197]:

df = df.set_index(keys='A', drop=False).reindex(np.arange(df.iloc[0]['A'], df.iloc[-1]['A'] + 0.5, 0.5))
df.reset_index(inplace=True) 
df['A'] = df['index']
df.drop(['A'], axis=1, inplace=True)
df.reset_index().drop(['level_0'], axis=1)
Out[197]:
   index   B   C
0    0.0   1   3
1    0.5   4   2
2    1.0   6   1
3    1.5 NaN NaN
4    2.0 NaN NaN
5    2.5 NaN NaN
6    3.0 NaN NaN
7    3.5   2   0
8    4.0   4   5
9    4.5   3   3

Here we set the index to column A but don't drop it and then reindex the df using the arange function.

Answer 4

This question was asked a long time ago, but I have a simple solution that's worth mentioning. You can simply use NumPy's NaN. For instance:

import numpy as np
df[i,j] = np.NaN

will do the trick.