检查字典中的唯一值并返回一个列表

Question

我一直在为这个练习挣扎几天，现在我发现每个近似值都有一个新问题，其思想是在字典上找到这些唯一值，并返回一个列表的钥匙

例如： 如果aDictionary = {1: 1, 3: 2, 6: 0, 7: 0, 8: 4, 10: 0}那么你的函数应该返回[1, 3, 8]，作为值1,2和4只出现一次。

这是我到目前为止已经试过：

def existsOnce(aDict): 

counting = {} 
tempList = [] 

for k in aDict.keys(): 
    print k, 
    print aDict[k] 


print 'values are:' 
for v in aDict.values(): 
    print v, 
    counting[v] = counting.get(v,0)+1 
    print counting[v] 
    tempNumbers = counting[v] 
    tempList.append(tempNumbers) 
print tempList 

如果我走这条路，我可以指出，并删除那些比一个大，但问题仍然存在，我将有一个零，我不想要它，因为在原始列表中不是唯一的。

def existsOnce2(aDict): 

# import Counter module in the top with `from collections import Counter` 

c = Counter() 

for letter in 'here is a sample of english text': 
    c[letter] += 1 
    if c[letter] == 1: 
     print c[letter],':',letter 

我试图用整数去检查哪些是从第一次出现，但不能将它翻译成字典或继续从这里出发。此外，我不确定导入模块是否允许在答案中，并且肯定必须是一种无需外部模块的方式。

def existsOnce3(aDict): 

    vals = {} 
    for i in aDict.values(): 
     for j in set(str(i)): 
      vals[j] = 1+ vals.get(j,0) 
    print vals 

    '''till here I get a counter of how many times a value appears in the original dictionary, now I should delete those bigger than 1''' 
    temp_vals = vals.copy() 
    for x in vals: 
     if vals[x] > 1: 
      print 'delete this: ', 'key:',x,'value:', vals[x] 
      temp_vals.pop(x) 
     else: 
      pass 
    print 'temporary dictionary values:', temp_vals 
    '''till here I reduced down the values that appear once, 1, 2 and 4, now I would need the go back and check the original dictionary and return the keys 
     Original dictionary: {1: 1, 3: 2, 6: 0, 7: 0, 8: 4, 10: 0} 
     temp_vals {'1': 1, '2': 1, '4': 1} 
     keys on temp_vals (1,2,4) are the values associated to the keys I got to retrieve from original dictionary (1,3,8) 
    ''' 
    print '---' 

    temp_list = [] 
    for eachTempVal in temp_vals: 
     temp_list.append(eachTempVal) 
    print 'temporary list values:', temp_list 
    ''' till here I got a temporary list with the values I need to search in aDict''' 
    print '---' 
    for eachListVal in temp_list: 
     print 'eachListVal:', eachListVal 
     for k,v in aDict.iteritems(): 
      print 'key:',k,'value:',v 

在这里，我可以不承担任何原因的值并加以比较，我已经试过类似语句来提取值：

if v == eachListVal: 
    do something 

但我做错了什么和不可以访问到价值。

Answer 1

你只需要使用您的vals字典，并保持键从aDict与有丘壑一个count == 1然后调用排序得到有序输出列表值：

def existsOnce3(aDict): 
    vals = {} 
    # create dict to sum all value counts 
    for i in aDict.values(): 
     vals.setdefault(i,0) 
     vals[i] += 1 
    # use each v/val from aDict as the key to vals 
    # keeping each k/key from aDict if the count is 1 
    return sorted(k for k, v in aDict.items() if vals[v] == 1) 

使用collections.Counter字典做计数只是调用计数器上的值，然后套用同样的逻辑，只要保持有AV计数== 1从柜台字典每个k：

from collections import Counter 
cn = Counter(aDict.values()) 
print(sorted(k for k,v in aDict.items() if cn[v] == 1)) 

Answer 2

如何：

from collections import Counter 

my_dict = {1: 1, 3: 2, 6: 0, 7: 0, 8: 4, 10: 0} 

val_counter = Counter(my_dict.itervalues()) 
my_list = [k for k, v in my_dict.iteritems() if val_counter[v] == 1] 

print my_list 

结果：

[1, 3, 8] 

Answer 3

一个衬里：

>>> aDictionary = {1: 1, 3: 2, 6: 0, 7: 0, 8: 4, 10: 0} 
>>> unique_values = [k for k,v in aDictionary.items() if list(aDictionary.values()).count(v)==1] 
>>> unique_values 
[1, 3, 8]