python统计文件行数和字符数组_使用Python查找文件大小的字符串中数组字符串的频率...

我看了很多答案，它们的目标是找到文件中每个单词的出现，或者大字符串，甚至数组。但我不想这么做，我的字符串也不是来自文本文件。在

给定一个大字符串，比如一个文件大小的字符串，如何计算大字符串中每个数组元素的频率(包括单词中的空格)？在def calculate_commonness(context, links):

c = Counter()

content = context.translate(string.maketrans("",""), string.punctuation).split(None)

for word in content:

if word in links:

c[word] += 1

print c

context = "It was November. Although it was November November November Passage not yet late, the sky was dark when I turned into Laundress Passage. Father had finished for the day, switched off the shop lights and closed the shutters; but so I would not come home to darkness he had left on the light over the stairs to the flat. Through the glass in the door it cast a foolscap rectangle of paleness onto the wet pavement, and it was while I was standing in that rectangle, about to turn my key in the door, that I first saw the letter. Another white rectangle, it was on the fifth step from the bottom, where I couldn\'t miss it."

links = ['November', 'Laundress', 'Passage', 'Father had']

# My output should look (something) like this:

# November = 4

# Laundress = 1

# Passage = 2

# Father had = 1

现在是十一月，洗衣店和通道，但不是'父亲有'。我需要能够找到带空格的字符串元素。我知道这是因为我将上下文拆分为“”返回“父”“had”，那么如何恰当地拆分上下文，还是将其与regex findall一起使用？在

编辑：

使用上下文作为一个大字符串，我有：

^{pr2}$

退货：Counter({'Laundress': 0, 'November': 0, 'Father had': 0, 'Passage': 0})