您可以这样做，我冒昧地重命名了列，以避免混淆我选择的内容，您可以使用以下命令对 Dataframe 执行相同操作：

data = data.rename(columns={0:'a',1:'b'})

In [41]:

data.merge(pd.DataFrame({'c':range(1,len(data[data.b.isnull()]) + 1)}, index=data[data.b.isnull()].index),how='left', left_index=True, right_index=True)
Out[41]:
     a    b   c
0  1.0  6.5 NaN
1  1.0  NaN   1
2  5.0  3.0 NaN
3  6.5  3.0 NaN
4  2.0  NaN   2

[5 rows x 3 columns]

下面是对这一行的一些解释：

# we want just the rows where column 'b' is null:
data[data.b.isnull()]

# now construct a dataset of the length of this dataframe starting from 1:
range(1,len(data[data.b.isnull()]) + 1) # note we have to add a 1 at the end

# construct a new dataframe from this and crucially use the index of the null values:
pd.DataFrame({'c':range(1,len(data[data.b.isnull()]) + 1)}, index=data[data.b.isnull()].index)

# now perform a merge and tell it we want to perform a left merge and use both sides indices, I've removed the verbose dataframe construction and replaced with new_df here but you get the point
data.merge(new_df,how='left', left_index=True, right_index=True)

编辑

你也可以用@Karl.D的建议换一种方式：

In [56]:

data['c'] = data['b'].isnull().cumsum().where(data['b'].isnull())
data
Out[56]:
     a    b   c
0  1.0  6.5 NaN
1  1.0  NaN   1
2  5.0  3.0 NaN
3  6.5  3.0 NaN
4  2.0  NaN   2

[5 rows x 3 columns]

计时还表明，卡尔的方法对于较大的数据集会更快，但我会对此进行描述：

In [57]:

%timeit data.merge(pd.DataFrame({'c':range(1,len(data[data.b.isnull()]) + 1)}, index=data[data.b.isnull()].index),how='left', left_index=True, right_index=True)
%timeit data['c'] = data['b'].isnull().cumsum().where(data['b'].isnull())
1000 loops, best of 3: 1.31 ms per loop
1000 loops, best of 3: 501 µs per loop

def function1(dd:pd.DataFrame):
    return dd.assign(col2=(dd['1'].isna()).cumsum()) if dd.name else dd
df1.groupby(df1['1'].isna()).apply(function1)

输出：

0    1  col2
0  1.0  6.5   NaN
1  1.0  NaN   1.0
2  5.0  3.0   NaN
3  6.5  3.0   NaN
4  2.0  NaN   2.0