''' 編寫(xiě)Python腳本，分析xx.log文件，按域名統(tǒng)計(jì)訪問(wèn)次數(shù)xx.log文件內(nèi)容如下： https://www.sogo.com/ale.html https://www.qq.com/3asd.html https://www.sogo.com/teoans.html https://www.bilibili.com/2 https://www.sogo.com/asd_sa.html https://y.qq.com/ https://www.bilibili.com/1 https://dig.chouti.com/ https://www.bilibili.com/imd.html https://www.bilibili.com/輸出： 4 www.bilibili.com 3 www.sogo.com 1 www.qq.com 1 y.qq.com 1 dig.chouti.com'''

首先我們拿到題目進(jìn)行需求分析：

1、先獲取數(shù)據(jù)就是域名??

獲取數(shù)據(jù)我們可以用正則，或者域名還是有相同點(diǎn)可以用split切分

2、統(tǒng)計(jì)域名訪問(wèn)的次數(shù)

可以用Python的內(nèi)置模塊來(lái)統(tǒng)計(jì)，

3、然后就是輸出要求的格式

sorted內(nèi)置函數(shù)用來(lái)排序

然后開(kāi)始最輕松的活，開(kāi)始碼字：

#第一種方式 import re from collections import Counter with open("xx.log","r",encoding="utf-8") as f:data=f.read()res=re.findall(r"https://(.*?)/.*?",data)dic=Counter(res)ret=sorted(dic.items(),key=lambda x:x[1],reverse=True)for k,v in ret:print(v,k)#第二種方式 dic={} with open("xx.log","r",encoding="utf-8") as f:for line in f:line=line.split("/")[2]if line not in dic:dic[line]=1else:dic[line]+=1 ret=sorted(dic.items(),key=lambda x:x[1],reverse=True) for k,v in ret:print( v,k)

這道題目考了這些知識(shí)點(diǎn)，re模塊，匿名函數(shù)，內(nèi)置函數(shù)sorted，collections中的Counter

這些在基礎(chǔ)篇都找得到相應(yīng)的博客，

我們就來(lái)說(shuō)說(shuō)collections中的Counter

我們直接打開(kāi)源碼

Counter類(lèi)的目的是用來(lái)跟蹤值出現(xiàn)的次數(shù)。它是一個(gè)無(wú)序的容器類(lèi)型，以字典的鍵值對(duì)形式存儲(chǔ)，其中元素作為key，其計(jì)數(shù)作為value。計(jì)數(shù)值可以是任意的Interger（包括0和負(fù)數(shù)）

再看源碼中的使用方法：

>>> c = Counter('abcdeabcdabcaba') # count elements from a string? ?生成計(jì)數(shù)對(duì)象

>>> c.most_common(3) # three most common elements? ?這里的3是找3個(gè)最常見(jiàn)的元素
[('a', 5), ('b', 4), ('c', 3)]

>>> c.most_common(4)? ? ??這里的4是找4個(gè)最常見(jiàn)的元素
[('a', 5), ('b', 4), ('c', 3), ('d', 2)]

>>> sorted(c) # list all unique elements? ?列出所有獨(dú)特的元素
['a', 'b', 'c', 'd', 'e']
>>> ''.join(sorted(c.elements())) # list elements with repetitions
'aaaaabbbbcccdde'

這里的elements 不知道是什么？那就繼續(xù)看源碼：

def elements(self):
　　'''Iterator over elements repeating each as many times as its count.

迭代器遍歷元素，每次重復(fù)的次數(shù)與計(jì)數(shù)相同

>>> sum(c.values()) # total of all counts? ?計(jì)數(shù)的總和
15

>>> c['a'] # count of letter 'a'? ??字母“a”的數(shù)

5

>>> for elem in 'shazam': # update counts from an iterable? 更新可迭代計(jì)數(shù)在新的可迭代對(duì)象
... c[elem] += 1 # by adding 1 to each element's count? ? ??在每個(gè)元素的計(jì)數(shù)中增加1
>>> c['a'] # now there are seven 'a'? ? ? ? ? ? 查看‘a(chǎn)’的計(jì)數(shù)，加上上面剛統(tǒng)計(jì)的2個(gè)，總共7個(gè)“a”
7
>>> del c['b'] # remove all 'b'? ? ? 刪除所有‘b’的計(jì)數(shù)
>>> c['b'] # now there are zero 'b'? ? ?
0

>>> d = Counter('simsalabim') # make another counter? ? ?
>>> c.update(d) # add in the second counter? ? ?在第二個(gè)計(jì)數(shù)器中添加

>>> c['a'] # now there are nine 'a'? ? ?
9

>>> c.clear() # empty the counter? ? qingg
>>> c
Counter()

Note: If a count is set to zero or reduced to zero, it will remain
in the counter until the entry is deleted or the counter is cleared:

如果計(jì)數(shù)被設(shè)置為零或減少到零，它將保持不變

在計(jì)數(shù)器中，直到條目被刪除或計(jì)數(shù)器被清除:

>>> c = Counter('aaabbc')
>>> c['b'] -= 2 # reduce the count of 'b' by two
>>> c.most_common() # 'b' is still in, but its count is zero
[('a', 3), ('c', 1), ('b', 0)]

大約就這幾個(gè)用法：大家拓展可以自己翻看源碼

轉(zhuǎn)載于:https://www.cnblogs.com/ManyQian/p/9152002.html

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