如何从 python 字典中删除一些值

Question

我有这个 python 字典myDict ：

{'Age': {0: '39'}, 'DailyRate': {0: '903'}, 'DistanceFromHome': {0: '2'}, 'EnvironmentSatisfaction': {0: '1'}, 'HourlyRate': {0: '41'}, 'JobInvolvement': {0: '4'}, 'JobLevel': {0: '3'}, 'JobSatisfaction': {0: '3'}, 'MonthlyIncome': {0: '7880'}, 'MonthlyRate': {0: '2560'}, 'NumCompaniesWorked': {0: '0'}, 'PercentSalaryHike': {0: '18'}, 'RelationshipSatisfaction': {0: '4'}, 'StandardHours': {0: '80'}, 'TotalWorkingYears': {0: '9'}, 'TrainingTimesLastYear': {0: '3'}, 'YearsAtCompany': {0: '8'}, 'YearsInCurrentRole': {0: '7'}, 'YearsSinceLastPromotion': {0: '0'}, 'YearsWithCurrManager': {0: '7'}, 'MaritalStatus_': {0: '2'}, 'JobRole_': {0: '7'}, 'Gender_': {0: '1'}, 'EducationField_': {0: '1'}, 'Department_': {0: '2'}, 'BusinessTravel_': {0: '2'}, 'OverTime_': {0: '1'}, 'Over18_': {0: '1'}}

如您所见，如果我从上面的示例中得到一个，如下所示，

{'Age': {0: '39'}}

值39前面多了一个0 。 这个零出现在每个键值对中。

我怎样才能摆脱这个0 ，所以它看起来像这样：

{'Age': '39'}

我试过这种方法，但它删除了整个密钥而不是0 ：

map(myDict.pop, ['Age',''])

有人可以帮帮我吗？

Answer 1

您可以使用字典理解来解决这个问题。 尝试做：

new_dict = {key: value[0] for key, value in old_dict.items()}

在这里，您遍历字典中的每个key, value对，并将新字典的键分配给旧字典的键。 但是该值成为字典内部字典的第0个键值。

例如， key从'Age'开始，因此新字典的第一个键是'Age' 。 然而，该值是{0: '39'}[0]即'39' 。 所以字典的第一个元素是'Age': '39'

Answer 2

您可以通过以下代码阅读它：

dict= {'Age': {0: '39'}, 'DailyRate': {0: '903'}, 'DistanceFromHome': {0: '2'}, 'EnvironmentSatisfaction': {0: '1'}, 'HourlyRate': {0: '41'}, 'JobInvolvement': {0: '4'}, 'JobLevel': {0: '3'}, 'JobSatisfaction': {0: '3'}, 'MonthlyIncome': {0: '7880'}, 'MonthlyRate': {0: '2560'}, 'NumCompaniesWorked': {0: '0'}, 'PercentSalaryHike': {0: '18'}, 'RelationshipSatisfaction': {0: '4'}, 'StandardHours': {0: '80'}, 'TotalWorkingYears': {0: '9'}, 'TrainingTimesLastYear': {0: '3'}, 'YearsAtCompany': {0: '8'}, 'YearsInCurrentRole': {0: '7'}, 'YearsSinceLastPromotion': {0: '0'}, 'YearsWithCurrManager': {0: '7'}, 'MaritalStatus_': {0: '2'}, 'JobRole_': {0: '7'}, 'Gender_': {0: '1'}, 'EducationField_': {0: '1'}, 'Department_': {0: '2'}, 'BusinessTravel_': {0: '2'}, 'OverTime_': {0: '1'}, 'Over18_': {0: '1'}}
print(dict['Age'][0])

转换只是做：

age = dict['Age'][0]
del(dict['Age'])
dict.update({"Age":age})

你将得到以下 dic 作为结果：

{'DailyRate': {0: '903'}, 'DistanceFromHome': {0: '2'}, 'EnvironmentSatisfaction': {0: '1'}, 'HourlyRate': {0: '41'}, 'JobInvolvement': {0: '4'}, 'JobLevel': {0: '3'}, 'JobSatisfaction': {0: '3'}, 'MonthlyIncome': {0: '7880'}, 'MonthlyRate': {0: '2560'}, 'NumCompaniesWorked': {0: '0'}, 'PercentSalaryHike': {0: '18'}, 'RelationshipSatisfaction': {0: '4'}, 'StandardHours': {0: '80'}, 'TotalWorkingYears': {0: '9'}, 'TrainingTimesLastYear': {0: '3'}, 'YearsAtCompany': {0: '8'}, 'YearsInCurrentRole': {0: '7'}, 'YearsSinceLastPromotion': {0: '0'}, 'YearsWithCurrManager': {0: '7'}, 'MaritalStatus_': {0: '2'}, 'JobRole_': {0: '7'}, 'Gender_': {0: '1'}, 'EducationField_': {0: '1'}, 'Department_': {0: '2'}, 'BusinessTravel_': {0: '2'}, 'OverTime_': {0: '1'}, 'Over18_': {0: '1'}, 'Age': '39'}

Answer 3

也许试试下面的代码：

keys = myDict.keys()
valuesWithZero = myDict.values()

valuesNoZero = []
for item in valuesWithZero:
    value_iterator = iter(item.values()) #to make dict_values obj iterable
    first_value = next(value_iterator) #obtaining first value
    valuesNoZero.append(first_value) #adding to new list
    

newDict = dict(zip(keys, valuesNoZero))  #combining keys arr and values arr
print(newDict)

# should output: {'Age': '39', 'DailyRate': '903', 'DistanceFromHome': '2', 'EnvironmentSatisfaction': '1', 'HourlyRate': '41', 'JobInvolvement': '4', 'JobLevel': '3', 'JobSatisfaction': '3', 'MonthlyIncome': '7880', 'MonthlyRate': '2560', 'NumCompaniesWorked': '0', 'PercentSalaryHike': '18', 'RelationshipSatisfaction': '4', 'StandardHours': '80', 'TotalWorkingYears': '9', 'TrainingTimesLastYear': '3', 'YearsAtCompany': '8', 'YearsInCurrentRole': '7', 'YearsSinceLastPromotion': '0', 'YearsWithCurrManager': '7', 'MaritalStatus_': '2', 'JobRole_': '7', 'Gender_': '1', 'EducationField_': '1', 'Department_': '2', 'BusinessTravel_': '2', 'OverTime_': '1', 'Over18_': '1'}

Answer 4

与接受的答案相同的过程，但使用了 Python 的一些功能特性。

import operator
zero = operator.itemgetter(0)

newdict = dict(zip(myDict,map(zero, myDict.values())))