Question

我有一个数据集，列中有一个名称列表，另一列中有每个名称的响应。每个名称都列出两次，我想看看两个记录的回复之间是否达成一致。即。

命名|回应1

命名|回应2

名称b |回应1

名称b |回应2

我创建了一个字典，其中键有两个值。字典创建名称作为键，每个响应作为值。我想创建一个列表来查看response1 = response2，或者response1！= response2。以下是我到目前为止的情况：

myDict = {}
if name not in myDict.keys():
    myDict[name] = {'response1': answer}
else:
    myDict[name]['reponse2'] = answer

match = True
for items in hospitalDict:
    if hospitalDict[items] != hospitalDict[items]:
          match = False
print match

我被困在这一部分...关于如何构建这个的任何建议？我还想最终将这些数据输出到csv。

Answer 1

我假设您想要实现的是具有匹配答案的名称列表。

鉴于您已创建字典：

myDict = {
    'name 1' : {'response1':'A', 'response2':'B'},
    'name 2' : {'response1':'C', 'response2':'C'},
    ... ,
    'name N' : {'response1':'Z', 'response2':'Z'},
}

您可以执行以下操作：

myList = []
for name, resp in myDict:
    if resp['response1'] == resp['response2']:
        myList.append(name)
print myList

结果应该是这样的：

['name 2', ..., 'name N']

该列表仅包含具有匹配答案的名称

Answer 2

我实现了NamesDict类，它可以为每个键（名称）提供2个或更多响应，比较它和导出。要在csv中导出，您可以轻松地将csv库用于python：https://docs.python.org/2/library/csv.html

我把比较和csv导出函数作为一个例子。

import csv


class NamesDict(dict):
    def __init__(self, *args, **kwargs):
        super(NamesDict, self).__init__(*args, **kwargs)

    def __setitem__(self, key, item):
        if isinstance(item, dict):
            self.__dict__[key] = item
        else:
            raise Exception('item has to be a dict')

    def __getitem__(self, key):
        return self.__dict__[key]

    def responses_match(self, key):
        # Here you can implement your own comparison method
        match = False
        for key_one, value_one in self.__dict__[key].items():
            for key_two, value_two in self.__dict__[key].items():
                if key_one != key_two and value_one == value_two:
                    match = True
        return match

    def export_csv(self, path):
        # Here you can change csv export
        with open(path, 'wb') as csv_file:
            fieldnames = ['name', 'responses']
            writer = csv.DictWriter(csv_file, fieldnames=fieldnames)
            for key, value in self.__dict__.items():
                responses_string = ''
                for resp_key, resp in value.items():
                    responses_string += '%s=%s\n' % (str(resp_key), str(resp_key))

                writer.writerow({
                    'name': key,
                    'responses': responses_string
                })


if __name__ == '__main__':
    namesDict = NamesDict()
    namesDict['test'] = {
        'response1': 1,
        'response2': 2
    }

    namesDict['test2'] = {
        'response1': 2,
        'response2': 2
    }

    print(namesDict.responses_match('test'))    # False
    print(namesDict.responses_match('test2'))   # True

    namesDict.export_csv('test.csv')

Answer 3

您可以使用groupby和customer key功能分隔两组

   from itertools import groupby
   myDict = {
        'name 1' : {'response1':'A', 'response2':'B'},
        'name 2' : {'response1':'C', 'response2':'C'},
        'name N' : {'response1':'Z', 'response2':'Z'},
    }

    for i in groupby(myDict.items(),key = lambda x: x[1]['response1'] == x[1]['response2']):
        print i[0],list(i[1])

    True [('name N', {'response2': 'Z', 'response1': 'Z'}), ('name 2', {'response2': 'C', 'response1': 'C'})]
    False [('name 1', {'response2': 'B', 'response1': 'A'})]

Answer 4

假设myDict和hospitalDict是相同的字典，您只需要一个简单的更改。改变这一行：

if hospitalDict[items] != hospitalDict[items]:

到线：

if hospitalDict[items]["response1"] != hospitalDict[items]["response2"]:

如果所有答案都匹配，匹配将为True，如果有一个或多个不同的答案，则为。

Answer 5

假设输入是以下形式的列表：

input_data = [
    [ "bob", "okay" ],
    [ "bob", "okay" ],
    [ "tom", "yes" ],
    [ "tom", "no" ],
    [ "kim", "red" ],
    [ "kim", "blue" ],
    [ "kim", "red" ],
    [ "kim", "green" ],
    [ "kim", "blue" ],
]

您可以按如下方式处理：

myDict = {}

for name, answer in input_data:
    myDict.setdefault(name, set()).add(answer)

print "The following names had differing responses"

for name, answers in myDict.items():
    if len(answers) > 1:
        print name, list(answers)

此代码段产生输出：

The following names had differing responses
kim ['blue', 'green', 'red']
tom ['yes', 'no']

Python Dictionary：比较1个键中的2个值

5 个答案: