如何将两列都与字符串列表进行比较并创建一个具有唯一项的新列? [英] How to compare two columns both with list of strings and create a new column with unique items?

问题描述

我有两列都带有字符串列表.基本上是一列 df['products'] 全部大写.另一列是产品描述df['desc'].

I have two columns both with list of strings. Basically one column df['products'] which are in all capitals. The other column is product description df['desc'].

我想检查 df['products'] 中的所有项目都存在于 df['desc'] 中，并从中创建一个新列.

I want to check what all items in df['products'] are present in df['desc'] and make a new column out of it.

我尝试了以下代码:

df['uniq'] = df.apply(lambda x : [i for i in x['products'] if i.lower() in x['desc']])

我检查了其他类似的问题并构建了上述代码，但它不起作用.

I checked the other similar questions and built the above code, but it's not working.

数据看起来像这样:

推荐答案

如果需要检查每行，似乎需要添加 axis=1 :

It seems you need add axis=1 if need check per rows:

df = pd.DataFrame({'products':[['A','B'],['D','C']],
                   'desc':[['a', 'c'],['c', 'e']]})

df['uniq'] = df.apply(lambda x: [i for i in x['products'] if i.lower() in x['desc']], axis=1)
print (df)
     desc products uniq
0  [a, c]   [A, B]  [A]
1  [c, e]   [D, C]  [C]

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