Question

我试图根据键列表中的值获取键，或者如果在字典中找不到值/键，则返回元素。

headersDict = {'Number; SEX AND AGE - Total Population':['TPop'],
               'Number; SEX AND AGE - Male Population':['MPop'],
               'Number; SEX AND AGE - Female Population':['FPop'],
               'Under 5 years': ['<5'],
               '5 to 9 years': ['5_9'],
               '10 to 14 years': ['10_14'],
               '15 to 19 years': ['15_19'],
               '20 to 24 years': ['20_24'],
               '25 to 29 years': ['25_29'],
               '30 to 34 years': ['30_34'],
               '35 to 39 years': ['35_39'],
               '40 to 44 years': ['40_44'],
               '45 to 49 years': ['45_49'],
               '50 to 54 years': ['50_54'],
               '55 to 59 years': ['55_59'],
               '60 to 64 years': ['60_64'],
               '65 to 69 years': ['65_69'],
               '70 to 74 years': ['70_74'],
               '75 to 79 years': ['75_79'],
               '80 to 84 years': ['80_84'],
               '85 years and over': ['85+'],
               'Median age(years)': ['Medage'],
               '16 years and over': ['16+'],
               '18 years and over': ['18+'],
               '21 years and over': ['21+'],
               '62 years and over': ['62+', 'sixty two+'],
               '65 years and over': ['65+', 'sixty five+']}

headersList = [  '1+', '25_29', '85+',
                '65+'
                ]
new_headersList = [k for k, v in headersDict.items() for elem in headersList for val in v if elem == val]


print(new_headersList)

如果尝试上述操作，则输出为：

$ python 1.py 
['25 to 29 years', '85 years and over', '65 years and over']

我需要的是：

$ python 1.py 
['1+', '25 to 29 years', '85 years and over', '65 years and over']

预先感谢您的帮助

Answer 1

此代码使字典反向，以便数组中的每个值成为新的键。使用该反向字典，可以轻松查询单个标题键或退回标题名称。

headersDict = {'Number; SEX AND AGE - Total Population': ['TPop'],
               'Number; SEX AND AGE - Male Population': ['MPop'],
               'Number; SEX AND AGE - Female Population': ['FPop'],
               'Under 5 years': ['<5'],
               '5 to 9 years': ['5_9'],
               '10 to 14 years': ['10_14'],
               '15 to 19 years': ['15_19'],
               '20 to 24 years': ['20_24'],
               '25 to 29 years': ['25_29'],
               '30 to 34 years': ['30_34'],
               '35 to 39 years': ['35_39'],
               '40 to 44 years': ['40_44'],
               '45 to 49 years': ['45_49'],
               '50 to 54 years': ['50_54'],
               '55 to 59 years': ['55_59'],
               '60 to 64 years': ['60_64'],
               '65 to 69 years': ['65_69'],
               '70 to 74 years': ['70_74'],
               '75 to 79 years': ['75_79'],
               '80 to 84 years': ['80_84'],
               '85 years and over': ['85+'],
               'Median age(years)': ['Medage'],
               '16 years and over': ['16+'],
               '18 years and over': ['18+'],
               '21 years and over': ['21+'],
               '62 years and over': ['62+', 'sixty two+'],
               '65 years and over': ['65+', 'sixty five+']}


headersDictReversed = {}
for k, v in headersDict.items():
  for new_k in v:
    headersDictReversed[new_k] = k

headersList = ['1+', '25_29', '85+', '65+']
results = []
for header in headersList:
  # Return the value for header and default to the header itself.
  results.append(headersDictReversed.get(header, header))
print(results)

['1 +'，'25至29岁'，'85岁以上'，'65岁以上']

Answer 2

如果可以使用熊猫，则可以使用以下解决方案：

import pandas as pd

df1 = pd.DataFrame(headersDict, index=[0,1]).T.reset_index()
df1 = pd.DataFrame(pd.concat([df1[0], df1[1]]).drop_duplicates()).join(df1, lsuffix='_1').drop(columns=['0',1]).rename(columns={'0_1':0}) 
a = pd.DataFrame(headersList).merge(df1, 'outer')[0:len(pd.DataFrame(headersList))].set_index(0)['index'] 
a.fillna(a.index.to_series()).values.tolist() 


# ['1+', '25 to 29 years', '85 years and over', '65 years and over']

根据字典列表中的值获取密钥

2 个答案: