字典键计数

Question

在Python 2.7中：我正在测量一个进程，该进程计算从函数返回的字典的键。

显示一个基本示例，其中函数 getList（）返回一个字符列表，可能是['a']，['b']，['c']或['d “];大多数列表是单个元素，但有时可能返回两个，例如['广告']。我想数一切都归来了。我想到这样做的方式如下所示：

    myDict = {'a':0, 'b':0, 'c':0, 'd':0, 'error':0, 'total':0}
    for key in charList:
        myDict[key] += 1
        myDict['total'] += 1

是否有更多的Pythonic方式，也许字典理解来计算列表中的键（长度不一样）？

import random

def getList():
    '''mimics a prcoess that returns a list of chars between a - d
       [most lists are single elements, though some are two elements]'''
    number = (random.randint(97,101))
    if number == 101:
        charList = [chr(number-1), chr(random.randint(97,100))]
        if charList[0] == charList[1]:
            getList()            
    else:
        charList = [chr(number)]
    return charList


myDict = {'a':0, 'b':0, 'c':0, 'd':0, 'error':0, 'total':0}

for counter in range(0,5):
    charList = getList()
    for key in charList:
        print charList, '\t', key
        try:
            myDict[key] += 1
            myDict['total'] += 1
        except:
            myDict['error'] += 1

print "\n",myDict

生成的输出：

Answer 1

您可以使用内置的collections.Counter课程：https://docs.python.org/2/library/collections.html#collections.Counter

例如使用您的代码：

import collections
ctr = collections.Counter()

for ii in range(0,5):
    charList = getList()
    ctr.update(charList)

ctr['total'] = sum(ctr.values())
print ctr

这将打印：

Counter({'total': 7, 'd': 5, 'a': 1, 'c': 1})

Answer 2

您可以使用collections.Counter：

# You need to initialize the counter or you won't get the entry with 0 count.
myDict = collections.Counter({'a': 0, 'b': 0, 'c': 0, 'd': 0})
myDict.update(x for _ in range(0, 5) for x in getList())
# Then create the 'total' entry
myDict['total'] = sum(myDict.values())

注意：如果'error'返回的list包含新字符getList()，'e'可能会向计数器添加新密钥而不设置'f'条目{1}}，...）。

Answer 3

使用collections.Counter和双循环生成器表达式将各个元素输入计数器：

>>> lst = [['a'], ['a', 'b'], ['c'], ['c', 'd']]
>>> c = collections.Counter((y for x in lst for y in x))
>>> c
Counter({'a': 2, 'c': 2, 'b': 1, 'd': 1})
>>> c.most_common(2)
[('a', 2), ('c', 2)]
>>> sum(c.values())
6

Answer 4

我能想到的最简单的方法是使用chain展开您的列表，然后使用Counter：让lst成为列表[['a'], ['b'], ['c'], ['d'], ['a', 'd']]

>>> from itertools import chain
>>> from collections import Counter
>>> c = Counter(chain(*lst))
>>> c['total'] = sum(c.values())
>>> c
Counter({'total': 6, 'd': 2, 'a': 2, 'b': 1, 'c': 1})