根据列的条件组合熊猫行（矢量化）

Question

我有这个数据框

import pandas as pd
df = pd.DataFrame({"a":[None, None, "hello1","hello2", None,"hello4","hello5","hello6", None, "hello8", None,"hello10",None ] , "b": ["we", "are the world", "we", "love", "the", "world", "so", "much", "and", "dance", "every", "day", "yeah"]})

    a   b
0   None    we
1   None    are the world
2   hello1  we
3   hello2  love
4   None    the
5   hello4  world
6   hello5  so
7   hello6  much
8   None    and
9   hello8  dance
10  None    every
11  hello10 day
12  None    yeah

所需的输出是：

    a       b       new_text
0   Intro   we      we are the world
2   hello1  we      we
3   hello2  love    love the
5   hello4  world   world
6   hello5  so      so
7   hello6  much    much and
9   hello8  dance   dance every
11  hello10 day     day yeah

我有一个函数可以做到这一点，但它在 Pandas 中使用，这可能不是最好的解决方案。

def connect_rows_on_condition(df, new_col_name, text, condition):
    if df[condition][0] == None:
        df[condition][0] = "Intro" 
    df[new_col_name] = ""
    index = 1
    last_non_none = 0
    while index < len(df):
        if df[condition][index] != None:
            last_non_none = index
            df[new_col_name][last_non_none] = df[text][index]

        elif df[condition][index] == None :
            df[new_col_name][last_non_none] = df[text][last_non_none] + " " + df[text][index]

        index += 1 

    output_df = df[df[condition].isna() == False]
    return output_df

主要逻辑是，如果“a”列中的“无”将 b 中的文本放入前一行。 有没有不基于循环的解决方案？

Answer 1

首先，创建一个描述组的系列：

grouping = df.a.notnull().cumsum()

然后，对于 a 列，我们可以使用第一个元素，对于 b 列，我们要连接所有元素：

df.groupby(grouping).agg({'a': 'first', 'b': ' '.join})

这给出：

         a                 b
a                           
0     None  we are the world
1   hello1                we
2   hello2          love the
3   hello4             world
4   hello5                so
5   hello6          much and
6   hello8       dance every
7  hello10          day yeah

如果需要，您可以将None替换为"Intro"作为特殊情况，因为该文本不会出现在输入中。

Answer 2

您还可以按单词进行分组，以防您没有空值可按此分组。 为了不重复约翰的解决方案，我将把它留在这里给那些可能对如何在非空情况下执行此操作感兴趣的人：

In [356]: Truth = pd.to_numeric(df.a.str.contains('None') == False).cumsum() 
     ...:                                                                                                                                                                                                  

In [357]: df.groupby(Truth)['b'].agg(list)                                                                                                                                                                 
Out[357]: 
a
0    [we, are the world]
1                   [we]
2            [love, the]
3                [world]
4                   [so]
5            [much, and]
6         [dance, every]
7            [day, yeah]
Name: b, dtype: object