Question

给出一个数据框

d = {'col1': [['how', 'are', 'you'], ['im', 'fine', 'thanks'], ['you', 'know'], [np.nan]],
     'col2': [['tell', 'how', 'me', 'you'], ['who', 'cares'], ['know', 'this', 'padewan'], ['who', 'are', 'you']]

df = pd.DataFrame(data=d)

我要创建第三列col3，它是col2列表中包含在col1列表中相应行的列表中的任何元素，否则{ {1}}。

必须采用任何匹配的元素。

在这种情况下，np.nan将是：

col3

我尝试过

           col1                      col2                           col3
0   ['how', 'are', 'you']      ['tell', 'how, 'me', 'you']        ['how', 'you']
1   ['im', 'fine', 'thanks']   ['who', 'cares']                   [np.nan] 
2   ['you', 'know']            ['know', 'this', 'padewan']        ['know']
3   [np.nan]                   ['who', 'are', 'you']              [np.nan]

这根本不起作用，所以任何想法都将非常有帮助。

Answer 1

类似这样的东西：

df['col3'] = [list(set(a).intersection(b)) for a, b in zip(df.col1, df.col2)]

输出：

                 col1                   col2        col3
0     [how, are, you]   [tell, how, me, you]  [you, how]
1  [im, fine, thanks]           [who, cares]          []
2         [you, know]  [know, this, padewan]      [know]
3               [nan]        [who, are, you]          []

Answer 2

另一个版本：

function extract(){
    var textsTol=GM_getValue("textsTol",[]);
    var texts = [];
    var list = document.querySelectorAll("div[class='r']")
    for (var i=0;i<list.length;i++){
        texts.push(list[i].textContent)


}
console.log(texts);
}

打印：

df['col3'] = df.apply(lambda x: [*set(x['col1']).intersection(x['col2'])] or [np.nan], axis=1 )

print(df)

Answer 3

我将在np.intersect1d的帮助下编写一个单独的函数并应用：

def intersect_nan(a,b):
    ret = np.intersect1d(a,b) 
    return list(ret) if len(ret)>0 else [np.nan]

df['col3'] = [intersect_nan(a,b) for a,b in zip(df['col1'], df['col2'])]

输出：

                 col1                   col2        col3
0     [how, are, you]   [tell, how, me, you]  [how, you]
1  [im, fine, thanks]           [who, cares]       [nan]
2         [you, know]  [know, this, padewan]      [know]
3               [nan]        [who, are, you]       [nan]

Answer 4

类似这样的东西：

 d =  {'col1': [['how', 'are', 'you'], ['im', 'fine', 'thanks'], ['you', 'know'], [numpy.nan]],
                'col2': [['tell', 'how', 'me', 'you'], ['who', 'cares'], ['know', 'this', 'padewan'],
                      ['who', 'are', 'you']]}
        df = pandas.DataFrame(d)
        list_col3 = []
        for index, row in df.iterrows():
            a_set= set(row['col1'])
            b_set= set(row['col2'])
            if len(a_set.intersection(b_set)) > 0:
                list_col3.append(list(a_set.intersection(b_set)))
            else:
                list_col3.append([numpy.nan])
        df['col3'] = list_col3
        print(df)

输出：

                 col1                   col2        col3
0     [how, are, you]   [tell, how, me, you]  [how, you]
1  [im, fine, thanks]           [who, cares]       [nan]
2         [you, know]  [know, this, padewan]      [know]
3               [nan]        [who, are, you]       [nan]

当在熊猫列中时如何检查一个列表中的元素是否为另一个

4 个答案: