在搜索列中搜索字符串并用字典键对它们进行分类

问题描述 投票:0回答:1

我已经导入了从LinkedIn导出的关于我的联系的电子表格,并希望将人们在不同级别上的职位进行分类。

因此,我创建了一个词典,其中包含用于查找每个职位级别的术语。

字典的第一个版本是:

dicpositions = {'0 - CEO, Founder': ['CEO', 'Founder', 'Co-Founder', 'Cofounder', 'Owner'],
                '1 - Director of': ['Director', 'Head'], 
                '2 - Manager': ['Manager', 'Administrador'], 
                '3 - Engenheiro': ['Engenheiro', 'Engineering'], 
                '4 - Consultor': ['Consultor', 'Consultant'], 
                '5 - Estagiário': ['Estagiário', 'Intern'], 
                '6 - Desempregado': ['Self-Employed', 'Autônomo'], 
                '7 - Professor': ['Professor', 'Researcher'] }

而且我需要一个代码来读取电子表格中的每个位置,检查是否有这些术语并在另一特定列中返回等效键。

我正在读取的数据帧的示例数据将是:

sample = pd.Series(data = (['(blank)'], ['Estagiário'], ['Professor', 'Adjunto'], 
                           ['CEO', 'and', 'Founder'], ['Engenheiro', 'de', 'Produção'], 
                           ['Consultant'], ['Founder', 'and', 'CTO'], 
                           ['Intern'], ['Manager', 'Specialist'], 
                           ['Administrador', 'de', 'Novos', 'Negócios'], 
                           ['Administrador', 'de', 'Serviços']))

哪个返回:

0                                [(blank)]
1                             [Estagiário]
2                     [Professor, Adjunto]
3                      [CEO, and, Founder]
4               [Engenheiro, de, Produção]
5                             [Consultant]
6                      [Founder, and, CTO]
7                                 [Intern]
8                    [Manager, Specialist]
9     [Administrador, de, Novos, Negócios]
10           [Administrador, de, Serviços]
dtype: object

我完成了以下代码:

import pandas as pd
plan = pd.read_excel('SpreadSheet Name.xlsx', sheet_name = 'Positions')

list0 = ['CEO', 'Founder', 'Co-Founder', 'Cofounder', 'Owner']
list1 = ['Director', 'Head']
list2 = ['Manager', 'Administrador']   
listgeral = [dic0, dic1, dic2]

def in_list(list_to_search,terms_to_search):     
    results = [item for item in list_to_search if item in terms_to_search]
    if len(results) > 0:
        return '0 - CEO, Founder'        
    else:
        pass
plan['PositionLevel'] = plan['Position'].str.split().apply(lambda x: in_list(x, listgeral[0]))

实际输出:

                                          Position           PositionLevel
0                                        '(blank)'                None
1                                     'Estagiário'                None
2                              'Professor Adjunto'                None
3                                'CEO and Founder'         '0 - CEO, Founder'
4                         'Engenheiro de produção'                None
5                                     'Consultant'                None
6                                'Founder and CTO'         '0 - CEO, Founder'
7                                         'Intern'                None
8                             'Manager Specialist'                None
9                'Administrador de Novos Negócios'                None

预期输出:

                                            Position         PositionLevel
0                                          '(blank)'              None
1                                       'Estagiário'       '5 - Estagiário'
2                                'Professor Adjunto'       '7 - Professor'
3                                  'CEO and Founder'      '0 - CEO, Founder'
4                           'Engenheiro de produção'       '3 - Engenheiro'
5                                       'Consultant'       '4 - Consultor'
6                                  'Founder and CTO'      '0 - CEO, Founder'
7                                           'Intern'       '5 - Estagiário'
8                               'Manager Specialist'        '2 - Manager'
9                  'Administrador de Novos Negócios'        '2 - Manager'

[首先,我打算为我的listgeral中的每个列表运行该代码,但我确实没有这样做。然后,我开始相信最好将它用于大型词典,就像从问题开头的dicpositions并返回术语的键一样。

我尝试将以下代码应用于该程序:

dictest = {'0 - CEO, Founder': ['CEO', 'Founder', 'Co-Founder', 'Cofounder', 'Owner'], 
           '1 - Director of': ['Director', 'Head'], 
           '2 - Manager': ['Manager', 'Administrador']}

def in_dic (x, dictest):
    for key in dictest:
        for elem in dictest[key]:
            if elem == x:
                return key
    return False

in_dic('CEO', dictest)的输出是'0 - CEO, Founder'

例如,in_dic('Banana', dictest)的输出为False

但是我无法从它开始并将此功能in_dic()应用到我的问题。

我真的很感谢任何人的帮助。

非常感谢。

我已经导入了从LinkedIn导出的关于我的联系的电子表格,并希望将人们在不同级别上的职位进行分类。因此,我创建了一个字典,其中包含用于查找每个词的术语...

python pandas dataframe dictionary series
1个回答
0
投票

我冒昧地重构了您的输入,但是这就是我所拥有的(可能有点过分设计)。简而言之,我们使用一个名为jellyfish的库(pip3 install jellyfish,代码取自this答案)进行模糊字符串匹配,以将excel工作表中的位置与dicpositions中的位置进行匹配,然后将它们映射到同一字典中的类别。这是导入和匹配功能:

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