Pandas：如何删除 Pandas 数据框中所有列的前导缺失值？

Question

With a pandas dataframe of the form:

     A     B     C
ID                
1   10   NaN   NaN
2   20   NaN   NaN
3   28  10.0   NaN
4   32  18.0  10.0
5   34  22.0  16.0
6   34  24.0  20.0
7   34  26.0  21.0
8   34  26.0  22.0

How can I remove a varying number of initial missing values? Initially, I'd like to forward fill the last values of the "new" columns so I'll end up with this:

    A     B     C
0  10  10.0  10.0
1  20  18.0  16.0
2  28  22.0  20.0
3  32  24.0  21.0
4  34  26.0  22.0
5  34  26.0  22.0
6  34  26.0  22.0
7  34  26.0  22.0

But I guess it would be just as natural to have nans on the remaining rows too:

    A     B     C
0  10  10.0  10.0
1  20  18.0  16.0
2  28  22.0  20.0
3  32  24.0  21.0
4  34  26.0  22.0
5  34  26.0   NaN
6  34   NaN   NaN
7  34   NaN   NaN

Here's a visual representation of the issue:

Before:

After:

I've come up with a cumbersome approach using a for loop where I remove the leading nans using df.dropna() , count the number of values I've removed (N), append the last available number N times, and build a new dataframe column by column. But this turned out to be pretty slow for larger dataframes. I feel like this is something that's already a built-in functionality of the omnipotent pandas library, but I haven't found anything so far. Does anyone have a suggestion to a less cumbersome way of doing this?

Complete code with a sample dataset:

import pandas as pd
import numpy as np

# sample dataframe
df = pd.DataFrame({'ID':[1,2,3,4,5,6,7,8],
                    'A': [10,20,28,32,34,34,34,34],
                   'B': [np.nan, np.nan, 10,18,22,24,26,26],
                    'C': [np.nan, np.nan, np.nan,10,16,20,21,22]})
df=df.set_index('ID')

# container for dataframe
# to be built using a for loop
df_new=pd.DataFrame()

for col in df.columns:
    # drop missing values column by column
    ser = df[col]
    original_length = len(ser)
    ser_new = ser.dropna()

    # if leading values are removed for N rows.
    # append last value N times for the last rows
    if len(ser_new) <= original_length:
        N = original_length - len(ser_new)
        ser_append = [ser.iloc[-1]]*N
        #ser_append = [np.nan]*N
        ser_new = ser_new.append(pd.Series(ser_append), ignore_index=True)
    df_new[col]=ser_new

df_new

Answer 1

We can make use of shift and move each series by the number of missing values

d = df.isna().sum(axis=0).to_dict() # calculate the number of missing rows per column 

for k,v in d.items():
    df[k] = df[k].shift(-v).ffill()

--

print(df)

   ID   A     B     C
0   1  10  10.0  10.0
1   2  20  18.0  16.0
2   3  28  22.0  20.0
3   4  32  24.0  21.0
4   5  34  26.0  22.0
5   6  34  26.0  22.0
6   7  34  26.0  22.0
7   8  34  26.0  22.0

Answer 2

Here is a pure Pandas solution. Use apply to shift the values up depending on number of leading NaN's and use ffill,

df.apply(lambda x: x.shift(-x.isna().sum())).ffill()


    A      B       C
1   10  10.0    10.0
2   20  18.0    16.0
3   28  22.0    20.0
4   32  24.0    21.0
5   34  26.0    22.0
6   34  26.0    22.0
7   34  26.0    22.0
8   34  26.0    22.0