如何打印字典的键和值作为表?

时间:2016-06-17 16:37:08

标签: python list csv dictionary

我有一个字典,用于存储每个键的项目列表,如下所示:

name_dict = {'MiddleName': ['H.', 'T.'], 'LastName': ['Perkins', 'Joseph'], 'FirstName': ['Elizabeth ', 'Scott ']}

我想以字典格式打印字典中的数据:

FirstName,MiddleName,LastName # the keys of the dictionary
Elizabeth,H.,Perkins #the values of the keys in one line (the first item in the list)
Scott,T.,Joseph #the values of the keys in new line (the following item in the list)

如何解决这个问题?

我已尝试通过Gareth Latty执行建议的解决方案,但这不起作用。

with open('C:/Output.csv',"w+") as file:
    w = csv.DictWriter(file,name_dict.keys())
    w.writeheader()
    w.writerow(name_dict) 

输出以下内容:

MiddleName,LastName,FirstName
"['H.', 'T.']","['Perkins', 'Joseph']","['Perkins', 'Joseph']"

知道如何输出新行中每个键的值(列表中的项目)吗?

5 个答案:

答案 0 :(得分:1)

我认为你滥用dict。如果您有多个值,则应使用list dicts而不是dict,其值为lists。而不是

Dict = {'MiddleName': ['H.', 'T.'], 'LastName': ['Perkins', 'Joseph'], 'FirstName': ['Elizabeth ', 'Scott ']}

你应该这样做:

Dict = [{'FirstName': 'Elizabeth', 'MiddleName': 'H.', 'LastName': 'Perkins'}, {'FirstName': 'Joseph', 'MiddleName': 'T. ', 'LastName': 'Scott'}]

或更易阅读的版本:

Dict = [
    {'FirstName': 'Elizabeth',   'MiddleName': 'H.',    'LastName': 'Perkins'},
    {'FirstName': 'Joseph',      'MiddleName': 'T. ',   'LastName': 'Scott'  }
]

如果你想打印一行(列表中的一个字典),你可以这样做:

def printDict(d):
    print d["FirstName"] + "," + d["MiddleName"] + "," + d["LastName"]

如果您想打印列表中的每个元素:

def printList(l):
    for i in l:
        printDict(i)

就这样使用它:

printList(Dict)

使用您的第一个(原始)Dict,访问Dict["FirstName"]将返回list,并且在打印时将打印为:

  

[" Elizabeth"," Joesph"]

但是使用第二种(我建议的新方式)Dict,访问Dict[0]["FirstName"]会返回string,并打印如下:

  

伊丽莎白

答案 1 :(得分:1)

csv.DictWriter期望每行有field:single_line对的字典,遗憾的是不是你拥有的字典,你基本上需要将你的数据结构转换为单行的dicts列表: / p>

[{'MiddleName': 'H.', 'FirstName': 'Elizabeth ', 'LastName': 'Perkins'}, {'MiddleName': 'T.', 'FirstName': 'Scott ', 'LastName': 'Joseph'}]

您可以使用以下内容进行转换:

import csv

def seperate_entries(dict_of_lists):    
    iters = [(k,iter(v)) for k,v in dict_of_lists.items()]
    try:
        while True:
            yield {k:next(v) for k,v in iters}
    except StopIteration:
        return
name_dict = {'MiddleName': ['H.', 'T.'], 'LastName': ['Perkins', 'Joseph'], 'FirstName': ['Elizabeth ', 'Scott ']}

with open('sample.csv',"w+") as file:
    w = csv.DictWriter(file,name_dict.keys())
    w.writeheader()
    w.writerows(seperate_entries(name_dict))

答案 2 :(得分:0)

要访问词典中的键,您只需执行以下操作:

middleNames=Dict['Middlename'] 
firstNames=Dict['FirstName']
lastNames=Dict['LastName']

您现在可以访问存储在内部列表信息中的值,这可以通过以下方式访问:

# Find how many items read (optional)
len(middleNames)

# Then iterate through the items
for mName in middleName:
   print mName   # this will print each list item e.g.

小时。

吨。

# for brevity, this is the same as doing this...
middleName[0] etc

希望这有帮助。

答案 3 :(得分:0)

我确信有更有效的方法可以达到你想要的效果,但这可以简单概述你想要的东西,并展示你如何实现它。

names = {'MiddleName': ['H.', 'T.'], 'LastName': ['Perkins', 'Joseph'], 'FirstName': ['Elizabeth ', 'Scott ']}

output = open('output.txt','w')

#NOTE: this will write headers in alphabetical format. You will need to change that so it follows the FirstName, MiddleName, LastName pattern
for key in sorted(names.keys()):
    output.write(key + ',')

output.write('\n')

#assuming we always have same amount of middle, first, and last names
for i in range(len(names['FirstName'])): 

    personFname = ''
    personLname = ''
    personMname = ''

    for key in names.keys():
        if key == 'MiddleName':
            personMname = names[key][i]
        elif key == 'FirstName':
            personFname = names[key][i]
        elif key == 'LastName':
            personLname = names[key][i]

    output.write(personFname + ',' + personMname + ',' + personLname)
    output.write('\n')

output.close()

答案 4 :(得分:0)

您需要定义您拥有的行数。 只需在Dict中将所有键平放到行中即可。

import csv

Dict = {'MiddleName': ['H.', 'T.'], 'LastName': ['Perkins', 'Joseph'], 'FirstName': ['Elizabeth ', 'Scott ']}

len_row = 2
with open('Output.csv', "w+") as file:
    w = csv.DictWriter(file, Dict.keys())
    w.writeheader()
    for i in range(len_row):
        row = {}
        for k in Dict.keys():
            for v in Dict.values():
                row[k] = Dict[k][i]
        w.writerow(row)
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