只是从上面使用了jiln的优秀答案，但需要展开以拆分多个列。我想分享一下。

def splitDataFrameList(df,target_column,separator):
''' df = dataframe to split,
target_column = the column containing the values to split
separator = the symbol used to perform the split

returns: a dataframe with each entry for the target column separated, with each element moved into a new row. 
The values in the other columns are duplicated across the newly divided rows.
'''
def splitListToRows(row, row_accumulator, target_columns, separator):
    split_rows = []
    for target_column in target_columns:
        split_rows.append(row[target_column].split(separator))
    # Seperate for multiple columns
    for i in range(len(split_rows[0])):
        new_row = row.to_dict()
        for j in range(len(split_rows)):
            new_row[target_columns[j]] = split_rows[j][i]
        row_accumulator.append(new_row)
new_rows = []
df.apply(splitListToRows,axis=1,args = (new_rows,target_column,separator))
new_df = pd.DataFrame(new_rows)
return new_df

这是一个相当直接的消息，它使用pandas str访问器的split方法，然后使用NumPy将每一行平铺成一个数组。

通过使用np.repeat以正确的次数重复未分割的列来检索相应的值。

var1 = df.var1.str.split(',', expand=True).values.ravel()
var2 = np.repeat(df.var2.values, len(var1) / len(df))

pd.DataFrame({'var1': var1,
              'var2': var2})

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

使用split(___， expand=True)和reset_index()的level和name参数的一行代码:

>>> b = a.var1.str.split(',', expand=True).set_index(a.var2).stack().reset_index(level=0, name='var1')
>>> b
   var2 var1
0     1    a
1     1    b
2     1    c
0     2    d
1     2    e
2     2    f

如果你想让b看起来和问题中的一模一样，你还可以这样做:

>>> b = b.reset_index(drop=True)[['var1', 'var2']]
>>> b
  var1  var2
0    a     1
1    b     1
2    c     1
3    d     2
4    e     2
5    f     2

我很欣赏“常舍”的回答，真的，但是iterrows()函数在大型数据集上花费很长时间。我面对了这个问题，然后我走到了这一步。

# First, reset_index to make the index a column
a = a.reset_index().rename(columns={'index':'duplicated_idx'})

# Get a longer series with exploded cells to rows
series = pd.DataFrame(a['var1'].str.split('/')
                      .tolist(), index=a.duplicated_idx).stack()

# New df from series and merge with the old one
b = series.reset_index([0, 'duplicated_idx'])
b = b.rename(columns={0:'var1'})

# Optional & Advanced: In case, there are other columns apart from var1 & var2
b.merge(
    a[a.columns.difference(['var1'])],
    on='duplicated_idx')

# Optional: Delete the "duplicated_index"'s column, and reorder columns
b = b[a.columns.difference(['duplicated_idx'])]

这样怎么样:

In [55]: pd.concat([Series(row['var2'], row['var1'].split(','))              
                    for _, row in a.iterrows()]).reset_index()
Out[55]: 
  index  0
0     a  1
1     b  1
2     c  1
3     d  2
4     e  2
5     f  2

然后你只需要重命名列

在添加了这个页面上所有解决方案中的一些零碎内容后，我能够得到这样的东西(对于需要立即使用它的人来说)。 函数的参数是df(输入数据帧)和key(用分隔符分隔字符串的列)。如果分隔符与分号“;”不同，只需替换为分隔符。

def split_df_rows_for_semicolon_separated_key(key, df):
    df=df.set_index(df.columns.drop(key,1).tolist())[key].str.split(';', expand=True).stack().reset_index().rename(columns={0:key}).loc[:, df.columns]
    df=df[df[key] != '']
    return df