我的问题与此类似，但答案似乎并不完全有效！
merge rows pandas dataframe based on condition
给出以下熊猫数据框：
+---------+-----------------+-----------------+
| SECTION | TEXT | NUMBER_OF_WORDS |
+---------+-----------------+-----------------+
| ONE | lots of text… | 55 |
+---------+-----------------+-----------------+
| ONE | word1 | 1 |
+---------+-----------------+-----------------+
| ONE | lots of text… | 151 |
+---------+-----------------+-----------------+
| ONE | word2 | 1 |
+---------+-----------------+-----------------+
| ONE | word3 | 1 |
+---------+-----------------+-----------------+
| ONE | word4 | 1 |
+---------+-----------------+-----------------+
| TWO | lots of text… | 523 |
+---------+-----------------+-----------------+
| TWO | lots of text… | 123 |
+---------+-----------------+-----------------+
| TWO | word4 | 1 |
+---------+-----------------+-----------------+
如果NUMBER_OF_WORDS列中的值为1；它必须与以上行合并；只要它们具有相同的SECTION值。
因此最终结果应如下所示：
+---------+--------------------------------------+-----------------+
| SECTION | TEXT | NUMBER_OF_WORDS |
+---------+--------------------------------------+-----------------+
| ONE | lots of text…, word1 | 56 |
+---------+--------------------------------------+-----------------+
| ONE | lots of text…, word2, word3, word4 | 154 |
+---------+--------------------------------------+-----------------+
| TWO | lots of text… | 523 |
+---------+--------------------------------------+-----------------+
| TWO | lots of text…, word4 | 124 |
+---------+--------------------------------------+-----------------+
这是代码；似乎可行，但并非如我所愿。
df.groupby(['SECTION', (df.NUMBER_OF_WORDS.shift(1) == 1)], as_index=False, sort=False).agg({'TEXT': lambda x: ', '.join(x), 'NUMBER_OF_WORDS': lambda x: sum(x)})
更新
这是BEN_YO的回答；但他似乎有轻微的错字。为了使该问题能够为将来的用户解答，我将略微修改他的回答。
s = df['NUMBER_OF_WORDS'].ne(1).cumsum()
out = df.groupby(s).agg({'SECTION': 'first','TEXT': lambda x: ', '.join(x),'NUMBER_OF_WORDS': lambda x: sum(x)})
答案 0 :(得分：1)
让我们尝试 System.out.println(count);
}
public WeakReference<A> m1() {
A a = new A();
return new WeakReference<>(a);
}
}
class A {
}
和groupby
cumsum