我导入了一个从Linkedin导出的电子表格，我想对不同级别的人的职位进行分类。
所以，我创建了一个字典，里面有查找每个职位级别的术语。
该词典的第一个版本是：
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'] }
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，并在另一个特定列中返回等效键。
我正在读取的数据帧的示例数据是：
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
这是我写作时得到的一系列列表：
plan['Position'].str.split()
我已经完成了以下代码：
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，我计划为listgeral中的每个列表运行该代码，但我做不到。然后我开始相信，将该代码应用于大词典会更好，就像问题开头的dicpositions和返回术语的关键字一样。
我已经尝试将以下代码应用于此程序：
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
其中输出来自：in_dic('CEO', dictest)
是'0 - CEO, Founder'
但我无法从中前进，并将函数in_dic()应用于我的问题。
我真的很感谢任何人的帮助。
非常感谢。
试试这个：
dicgeral_2 = {'0 - Teste': ['Accelerator', 'Account'], '1 - Teste': ['Académico'], '2 - Teste': ['Visual', 'Volunteer']}
def in_dict(pos, dic):
for key, value in dic.items():
if pos in value:
return key
return False
in_dict("Account", dicgeral_2)
输出：
>>0 - Teste