Python中的dict（）有问题，TypeError:“tuple”对象不是callab

Question

我正在尝试创建一个函数，它将接受任意数量的字典输入，并创建一个包含所有输入的新字典。如果两个键相同，则该值应为一个列表，其中包含两个值。我已经成功地做到了这一点——但是，dict（）函数有问题。如果我在python shell中手动执行dict函数，我就可以创建一个新的字典而不会有任何问题；但是，当它嵌入到我的函数中时，我会得到一个TypeError。下面是我的代码：

#Module 6 Written Homework
#Problem 4

dict1= {'Fred':'555-1231','Andy':'555-1195','Sue':'555-2193'}
dict2= {'Fred':'555-1234','John':'555-3195','Karen':'555-2793'}

def dictcomb(*dict):
    mykeys = []
    myvalues = []
    tupl = ()
    tuplist = []
    newtlist = []
    count = 0
    for i in dict:
        mykeys.append(list(i.keys()))
        myvalues.append(list(i.values()))
        dictlen = len(i)
        count = count + 1
    for y in range(count):
        for z in range(dictlen):
            tuplist.append((mykeys[y][z],myvalues[y][z]))
    tuplist.sort()
    for a in range(len(tuplist)):
        try:
            if tuplist[a][0]==tuplist[a+1][0]:
                comblist = [tuplist[a][1],tuplist[a+1][1]]
                newtlist.append(tuple([tuplist[a][0],comblist]))
                del(tuplist[a+1])
            else:
                newtlist.append(tuplist[a])
        except IndexError as msg:
            pass
    print(newtlist)
    dict(newtlist)

我得到的错误如下：

Traceback (most recent call last):
  File "<pyshell#17>", line 1, in <module>
    dictcomb(dict1,dict2)
  File "C:\Python33\M6HW4.py", line 34, in dictcomb
    dict(newtlist)
TypeError: 'tuple' object is not callable

如上所述，在python shell中，print（newtlist）给出：

[('Andy', '555-1195'), ('Fred', ['555-1231', '555-1234']), ('John', '555-3195'),     ('Karen', '555-2793')]

如果将此输出复制并粘贴到dict（）函数中：

dict([('Andy', '555-1195'), ('Fred', ['555-1231', '555-1234']), ('John', '555-3195'), ('Karen', '555-2793')])

输出变成我想要的，即：

{'Karen': '555-2793', 'Andy': '555-1195', 'Fred': ['555-1231', '555-1234'], 'John': '555-3195'}

不管我做什么，我都不能在我的函数中复制这个。请帮帮我！谢谢您！

Answer 1

不应将关键字用作变量名的典型示例。这里dict(newtlist)试图调用dict()内置python，但是有一个冲突的局部变量dict。重命名该变量以解决问题。

像这样的：

def dictcomb(*dct): #changed the local variable dict to dct and its references henceforth
    mykeys = []
    myvalues = []
    tupl = ()
    tuplist = []
    newtlist = []
    count = 0
    for i in dct:
        mykeys.append(list(i.keys()))
        myvalues.append(list(i.values()))
        dictlen = len(i)
        count = count + 1
    for y in range(count):
        for z in range(dictlen):
            tuplist.append((mykeys[y][z],myvalues[y][z]))
    tuplist.sort()
    for a in range(len(tuplist)):
        try:
            if tuplist[a][0]==tuplist[a+1][0]:
                comblist = [tuplist[a][1],tuplist[a+1][1]]
                newtlist.append(tuple([tuplist[a][0],comblist]))
                del(tuplist[a+1])
            else:
                newtlist.append(tuplist[a])
        except IndexError as msg:
            pass
    print(newtlist)
    dict(newtlist)

Answer 2

函数有一个名为dict的局部变量，它来自函数参数并屏蔽内置的dict()函数：

def dictcomb(*dict):
              ^
            change to something else, (*args is the typical name)

Answer 3

您需要自己完全实现这个功能，还是可以使用defaultdict？如果是这样的话，你可能会做如下事情：

from collections import defaultdict

merged_collection = defaultdict(list)
collection_1= {'Fred':'555-1231','Andy':'555-1195','Sue':'555-2193'}
collection_2= {'Fred':'555-1234','John':'555-3195','Karen':'555-2793'}

for collection in (collection_1, collection_2):
    for name, number in collection.items():
        merged_collection[name].append(number)

for name, number in merged_collection.items(): 
    print(name, number)

John ['555-3195']
Andy ['555-1195']
Fred ['555-1231', '555-1234']
Sue ['555-2193']
Karen ['555-2793']