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python更新列表中的dict值

[英]python to update dict values in list

I have a list with two element (key-value pair)in it. 我有一个包含两个元素(键值对)的列表。 Both element has same keys in it. 两个元素都有相同的键。 however few values are different, shown as below. 但是很少有不同的值,如下所示。

alist = [{u'a': u'x', u'b': u'y', u'c': u'z'}, {u'a': u'x', u'b': u'm', u'c': u'n'}]

I want to update new data to alist from available in dictionary, which is update from user input, shown below. 我想更新新数据,从字典中可用,这是从用户输入更新,如下所示。

a_dict  = {'a': ['user_input_x'], 'b': ['user_input_y', 'user_input_m'], 'c': ['user_input_z', 'user_input_n']}

Resultant alist should look like this, 结果alist应该是这样的,

ans_alist = [{u'a': u'user_input_x', u'b': u'user_input_y', u'c': u'user_input_z'}, {u'a': u'user_input_x', u'b': u'user_input_m', u'c': u'user_input_n'}] 

I was trying out few things, below is the snippet, however since it's dict, code is updating all the keys, with same values, 我正在尝试一些事情,下面是片段,但是因为它是dict,代码正在更新所有键,具有相同的值,

for i in range(0, 2):
    for key in alist:
        key['a'] = a_dict['a'][i]
        key['b'] = a_dict['b'][i]
        key['c'] = a_dict['b'][i]
print alist


ans_alist = [{u'a': ['user_input_x'], u'b': 'user_input_y', u'c': 'user_input_n'}, {u'a': ['user_input_x'], u'b': 'user_input_y', u'c': 'user_input_n'}]

appreciate your help. 感谢您的帮助。

New Answer : 新答案

Update alist according to the values in a_dict . 根据a_dict的值更新alist

>>> new_alist = []
>>> for i,d in enumerate(alist):                   #traverse the list
       temp = {} 

       for key,val in d.items():                   #go through the `dict`
           if len(a_dict[key])>0 :                 #alist->[key] is present in a_dict
               temp[key] = a_dict[key].pop(0)      #used the value, delete it from a_dict
           else :                                  #use previous updated value
               temp[key] = new_alist[i-1][key]     #get value from previously updated 

       new_alist.append(temp)                      #add the updated `dict`     

#driver values #driver值

IN :  alist = [{u'a': u'x', u'b': u'y', u'c': u'z'}, 
               {u'a': u'x', u'b': u'm', u'c': u'n'}]
IN :  a_dict  = {'a': ['user_input_x'], 'b': ['user_input_y', 'user_input_m'], 'c': ['user_input_z', 'user_input_n']}

OUT : new_alist
     => [{'a': 'user_input_x', 'b': 'user_input_y', 'c': 'user_input_z'}, 
         {'a': 'user_input_x', 'b': 'user_input_m', 'c': 'user_input_n'}]

Old Answer : (wasn't according to requirements fully) 旧答案 :(完全不符合要求)

Compute a_dict : 计算a_dict

>>> from collections import defaultdict
>>> a_dict = defaultdict(list)

>>> for d in alist:                                         #traverse the list
        for key,val in d.items(): 
            if val not in a_dict[key]:                      #if val not already there
               a_dict[key].append(val)                      #add it to the [key]

>>> a_dict
=> defaultdict(<class 'list'>, 
       {'a': ['user_input_x'], 
        'b': ['user_input_y', 'user_input_m'], 
        'c': ['user_input_z', 'user_input_n']
   })

Compute ans_list : 计算ans_list

>>> ans_list = []
>>> for d in alist:                                         #traverse the list
        temp = {} 
        for key,val in d.items(): 
            temp[key] = val 
        ans_list.append(temp)                               #add the new dictionary

>>> ans_list
=> [{'a': 'user_input_x', 'b': 'user_input_y', 'c': 'user_input_z'}, 
    {'a': 'user_input_x', 'b': 'user_input_m', 'c': 'user_input_n'}]

Another alternative way would be: 另一种替代方式是:

from collections import defaultdict
repeat_count = len(alist)
ans_alist = [defaultdict(dict) for _ in range(repeat_count)]

for k, v in a_dict.items():
    mul_times = repeat_count / len(v)   # Use // if Python 3
    extend_counter = repeat_count % (mul_times * len(v))
    repeat_v = v * mul_times + v[:extend_counter]

    unicode_k = u'{}'.format(k)
    for index, val in enumerate(repeat_v):
        ans_alist[index][unicode_k] = u'{}'.format(val)

You need to iterate over the list of dicts, grab one dict at a time. 你需要遍历dicts列表,一次抓取一个dict。 Then iterate over its keys and pop the items from the result list: 然后遍历其键并弹出结果列表中的项:

new_alist = []
for d in alist:
    new_dict = {}
    for k in d.keys():
        if a_dict[k]:
            new_dict[k] = a_dict[k].pop(0)
    new_alist.append(new_dict)

>>> new_alist
[{u'a': 'user_input_x', u'c': 'user_input_z', u'b': 'user_input_y'}, {u'c': 
'user_input_n', u'b': 'user_input_m'}]

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