一种使用.apply()改变列格式的简单方法，以便.explod()可以使用它:

import string
import pandas as pd
from io import StringIO

file = StringIO("""    var1  var2
0  a,b,c     1
1  d,e,f     2""")

df = pd.read_csv(file, sep=r'\s\s+')

df['var1'] = df['var1'].apply(lambda x : str(x).split(','))

df.explode('var1')

输出:

  var1  var2
0   a   1
0   b   1
0   c   1
1   d   2
1   e   2
1   f   2

基于优秀的@DMulligan的解决方案，这里有一个通用的向量化(无循环)函数，它将数据帧的一列分割成多行，并将其合并回原始数据帧。它还从这个答案中使用了一个很棒的通用change_column_order函数。

def change_column_order(df, col_name, index):
    cols = df.columns.tolist()
    cols.remove(col_name)
    cols.insert(index, col_name)
    return df[cols]

def split_df(dataframe, col_name, sep):
    orig_col_index = dataframe.columns.tolist().index(col_name)
    orig_index_name = dataframe.index.name
    orig_columns = dataframe.columns
    dataframe = dataframe.reset_index()  # we need a natural 0-based index for proper merge
    index_col_name = (set(dataframe.columns) - set(orig_columns)).pop()
    df_split = pd.DataFrame(
        pd.DataFrame(dataframe[col_name].str.split(sep).tolist())
        .stack().reset_index(level=1, drop=1), columns=[col_name])
    df = dataframe.drop(col_name, axis=1)
    df = pd.merge(df, df_split, left_index=True, right_index=True, how='inner')
    df = df.set_index(index_col_name)
    df.index.name = orig_index_name
    # merge adds the column to the last place, so we need to move it back
    return change_column_order(df, col_name, orig_col_index)

例子:

df = pd.DataFrame([['a:b', 1, 4], ['c:d', 2, 5], ['e:f:g:h', 3, 6]], 
                  columns=['Name', 'A', 'B'], index=[10, 12, 13])
df
        Name    A   B
    10   a:b     1   4
    12   c:d     2   5
    13   e:f:g:h 3   6

split_df(df, 'Name', ':')
    Name    A   B
10   a       1   4
10   b       1   4
12   c       2   5
12   d       2   5
13   e       3   6
13   f       3   6    
13   g       3   6    
13   h       3   6    

注意，它保留了列的原始索引和顺序。它也适用于具有非连续索引的数据框架。

有可能在不改变数据框架结构的情况下拆分和爆炸数据框架

拆分和展开特定列的数据

输入:

    var1    var2
0   a,b,c   1
1   d,e,f   2



#Get the indexes which are repetative with the split 
df['var1'] = df['var1'].str.split(',')
df = df.explode('var1')

Out:

    var1    var2
0   a   1
0   b   1
0   c   1
1   d   2
1   e   2
1   f   2

Edit-1

对多列的行进行拆分和展开

Filename    RGB                                             RGB_type
0   A   [[0, 1650, 6, 39], [0, 1691, 1, 59], [50, 1402...   [r, g, b]
1   B   [[0, 1423, 16, 38], [0, 1445, 16, 46], [0, 141...   [r, g, b]

基于参考列重新索引，并将列值信息与堆栈对齐

df = df.reindex(df.index.repeat(df['RGB_type'].apply(len)))
df = df.groupby('Filename').apply(lambda x:x.apply(lambda y: pd.Series(y.iloc[0])))
df.reset_index(drop=True).ffill()

Out:

                Filename    RGB_type    Top 1 colour    Top 1 frequency Top 2 colour    Top 2 frequency
    Filename                            
 A  0       A   r   0   1650    6   39
    1       A   g   0   1691    1   59
    2       A   b   50  1402    49  187
 B  0       B   r   0   1423    16  38
    1       B   g   0   1445    16  46
    2       B   b   0   1419    16  39

升级了MaxU的答案，支持MultiIndex

def explode(df, lst_cols, fill_value='', preserve_index=False):
    """
    usage:
        In [134]: df
        Out[134]:
           aaa  myid        num          text
        0   10     1  [1, 2, 3]  [aa, bb, cc]
        1   11     2         []            []
        2   12     3     [1, 2]      [cc, dd]
        3   13     4         []            []

        In [135]: explode(df, ['num','text'], fill_value='')
        Out[135]:
           aaa  myid num text
        0   10     1   1   aa
        1   10     1   2   bb
        2   10     1   3   cc
        3   11     2
        4   12     3   1   cc
        5   12     3   2   dd
        6   13     4
    """
    # make sure `lst_cols` is list-alike
    if (lst_cols is not None
        and len(lst_cols) > 0
        and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
        lst_cols = [lst_cols]
    # all columns except `lst_cols`
    idx_cols = df.columns.difference(lst_cols)
    # calculate lengths of lists
    lens = df[lst_cols[0]].str.len()
    # preserve original index values    
    idx = np.repeat(df.index.values, lens)
    res = (pd.DataFrame({
                col:np.repeat(df[col].values, lens)
                for col in idx_cols},
                index=idx)
             .assign(**{col:np.concatenate(df.loc[lens>0, col].values)
                            for col in lst_cols}))
    # append those rows that have empty lists
    if (lens == 0).any():
        # at least one list in cells is empty
        res = (res.append(df.loc[lens==0, idx_cols], sort=False)
                  .fillna(fill_value))
    # revert the original index order
    res = res.sort_index()
    # reset index if requested
    if not preserve_index:        
        res = res.reset_index(drop=True)

    # if original index is MultiIndex build the dataframe from the multiindex
    # create "exploded" DF
    if isinstance(df.index, pd.MultiIndex):
        res = res.reindex(
            index=pd.MultiIndex.from_tuples(
                res.index,
                names=['number', 'color']
            )
    )
    return res

我有一个类似的问题，我的解决方案是将数据帧转换为字典列表，然后进行转换。函数如下:

import re
import pandas as pd

def separate_row(df, column_name):
    ls = []
    for row_dict in df.to_dict('records'):
        for word in re.split(',', row_dict[column_name]):
            row = row_dict.copy()
            row[column_name]=word
            ls.append(row)
    return pd.DataFrame(ls)

例子:

>>> from pandas import DataFrame
>>> import numpy as np
>>> a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
               {'var1': 'd,e,f', 'var2': 2}])
>>> a
    var1  var2
0  a,b,c     1
1  d,e,f     2
>>> separate_row(a, "var1")
  var1  var2
0    a     1
1    b     1
2    c     1
3    d     2
4    e     2
5    f     2

您还可以稍微更改该函数以支持分离列表类型行。

Try:

vals = np.array(a.var1.str.split(",").values.tolist())    
var = np.repeat(a.var2, vals.shape[1])

out = pd.DataFrame(np.column_stack((var, vals.ravel())), columns=a.columns)
display(out)

      var1 var2
    0   1   a
    1   1   b
    2   1   c
    3   2   d
    4   2   e
    5   2   f