Python 将字符串和整数列相乘
我有一个具有这种结构[1]的数据帧,我想将字符串和整数列相乘Python 将字符串和整数列相乘,python,pandas,dataframe,Python,Pandas,Dataframe,我有一个具有这种结构[1]的数据帧,我想将字符串和整数列相乘 +----------------------+------------+-------------------------+-----------+--+ | url | date | word | mentioned | | |----------------------+------------+------------------------
+----------------------+------------+-------------------------+-----------+--+
| url | date | word | mentioned | |
|----------------------+------------+-------------------------+-----------+--+
| newspaperarticle.com | 2018-12-22 | [canada,house,micheal] | [2,2,1] | |
| articleUSA.com | 2018-12-23 | [new york,murder,angry] | [2,3,1] | |
+----------------------+------------+-------------------------+-----------+-
我要在列名中乘以单词数
+----------------------+------------+-------------------------+-------+---+--+
| url | date | word |mentioned
|----------------------+------------+-------------------------+-------+---+--+
| newspaperarticle.com | 2018-12-22 | [canada,canada,house,..] |[2,2,1]
| articleUSA.com | 2018-12-23 | [new york,new york,murder,..] |[2,3,1]
+----------------------+------------+-------------------------+-------+---+--+
到目前为止,我所做的是用不起作用的乘法方法将列相乘。我还尝试了for循环,对单个元素进行索引并将它们相乘,但总是将错误字符串从索引中删除。您可以使用级别为0的聚合为列表:
s = [df[i].explode() for i in ['word','mentioned']]
df['word'] = s[0].repeat(s[1]).groupby(level=0).agg(list)
注意:这是假设所提到的
word
和列是一系列列表,而不是列表的字符串表示形式。ast.literal\u eval
将在字符串[canada,canada,house]
上失败,因为缺少字符串分隔符。我真诚地希望这些是列表。@Błotosmętek你是对的,删除了那张便条,然后我们必须处理输入数据。我也希望如此:)不用担心,这些都是列表,谢谢你从未听说过重复函数。这种方法比执行多个for循环和围绕索引工作要容易得多:)
print(df)
url date \
0 newspaperarticle.com 2018-12-22
1 articleUSA.com 2018-12-23
word mentioned
0 [canada, canada, house, house, micheal] [2, 2, 1]
1 [new york, new york, murder, murder, murder, a... [2, 3, 1]