Question

输入：

我有一个如下所示的数据框

Job job = Job.getInstance();

输出：

我需要找到Remaining_name列，如下所示

Full_Name                     Name1          Name2 
John Mathew Davidson          John           Davidson
Paul Theodre Luther           Paul           Theodre
Victor George Mary            George         Mary

说明：

我需要在另一列的值（句子）中比较一个以上的列的值（单词），并找到可能在整个字符串的任何位置出现的不匹配单词。

Answer 1

使用替换的矢量化解决方案

df['Remaining_name'] = df.apply(lambda x: x['Full_Name'].replace(x['Name1'], '').replace(x['Name2'], ''), axis=1).str.strip()


    Full_Name               Name1   Name2       Remaining_name
0   John Mathew Davidson    John    Davidson    Mathew
1   Paul Theodre Luther     Paul    Theodre     Luther
2   Victor George Mary      George  Mary        Victor

编辑：如果有很多以“名称”开头的列，则可以选择一个切片并根据正则表达式模式替换Full_Name中的值

df['tmp'] = df[df.columns[df.columns.str.startswith('Name')]].apply('|'.join, axis = 1)
df['Remaining_name'] = df.apply(lambda x: x.replace(x['tmp'], '', regex = True), axis = 1)['Full_Name'].str.strip()
df.drop('tmp', axis =1, inplace = True)


    Full_Name                   Name1   Name2       Remaining_name
0   John Mathew Davidson        John    Davidson    Mathew
1   Paul Theodre Luther         Paul    Theodre     Luther
2   Victor George Mary          George  Mary        Victor
3   Henry Patrick John Harrison Henry   John        Patrick Harrison

Answer 2

这是您提供的数据：

import pandas as pd

full_name = ['John Mathew Davidson', 'Paul Theodre Luther', 'Victor George Mary']
name_1 = ['John', 'Paul', 'George']
name_2 = ['Davidson', 'Theodre', 'Mary']

df = pd.DataFrame({'Full_Name':full_name, 'Name1':name_1, 'Name2':name_2 })

为了对一行中的多个列执行操作，最好的办法是分别定义该函数。它使代码更具可读性，更易于调试该函数将以DataFrame行作为输入：

def find_missing_name(row):

    known_names = [row['Name1'], row['Name2']] ## we add known names to a list to check it later    

    full_name_list = row['Full_Name'].split(' ') ## converting the full name to the list by splitting it on spaces

    ## WARNING! this function works well only if you are sure your 'Full_Name' column items are separated by a space.

    missing_name = [x for x in full_name_list if x not in known_names] ## looping throught the full name list and comparing it to the known_names list, to only keep the missing ones.
    missing_name = ','.join(missing_name) ## in case there are more than one missing names convert them all in a string separated by comma

    return missing_name

现在将功能应用于现有的DataFrame：

df['missing_name'] = df.apply(find_missing_name, axis=1) ## axis=1 means 'apply to each row', where axis=0 means 'apply to each column'

输出：

希望这会有所帮助:)

Answer 3

您可以使用以下代码一行完成操作：

df['Remaining_name'] = df.apply(lambda x: [i for i in x['Full_Name'].split() if all(i not in x[c] for c in df.columns[1:])], axis=1)

这将以Remaining_name的形式返回list列，但是在您的名称包含三个以上子字符串的情况下，此功能将很有用，例如：

                     Full_Name    Name1     Name2    Remaining_name
0         John Mathew Davidson     John  Davidson          [Mathew]
1          Paul Theodre Luther     Paul   Theodre          [Luther]
2           Victor George Mary   George      Mary          [Victor]
3  Henry Patrick John Harrison  Patrick     Henry  [John, Harrison]

Answer 4

尝试一下：

import numpy as np

In [835]: df
Out[835]: 
              Full_name   Name1     Name2
0  John Mathew Davidson    John  Davidson
1   Paul Theodre Luther    Paul   Theodre
2    Victor George Mary  George      Mary

ll = []

In [854]: for i, r in df.iterrows():
     ...:     big_list = r[0].split(' ')
     ...:     l1 = [r[1]]
     ...:     l2 = [r[2]]
     ...:     remaining_item = np.setdiff1d(big_list, l1+l2)[0]
     ...:     ll.append(remaining_item)
In [856]: df['Remaining_name'] = ll

In [857]: df
Out[857]: 
              Full_name   Name1     Name2 Remaining_name
0  John Mathew Davidson    John  Davidson         Mathew
1   Paul Theodre Luther    Paul   Theodre         Luther
2    Victor George Mary  George      Mary         Victor

熊猫-通过比较同一数据框中的其他列值（单词）来提取一列中不匹配的单词

4 个答案: