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python常见问题解决方案

2016-12-08 10:44 155 查看
平时工作中经常需要用到这些python小技巧,顺便做个记录


import requests
import time

def get_pr(domain):
pr = 6
time.sleep(1)
html = requests.get("http://pr.web3389.com/pr/%s" % domain.strip())
string = html.text
key = "images/pr1/Rank_"
ipos = string.find(key)
pr = string[ipos+ len(key):]
pr = pr[:1]
try:
if int(pr) < 5:
pr = 5
except:
pr = 5
n = int(pr) - 4
tmp = 10 * 10**(n)
count = tmp + count

with open('domain_date.txt') as f:
for line in f.readlines():
url = line.split(' ')[0]
month = line.split(' ')[-1].split('-')[1]
try:
pr = get_pr(url)
except:
pass
print(pr,url,month)

fuck = '{}:{}:{}'.format(url,month,pr)
with open('new_domain.txt','a') as file:
file.writelines(fuck + ‘\n’)


  

对字典value进行排序
import operator

diaosi = {}
with open('all.txt', 'r', encoding="utf-8") as f:
for line in f.readlines():
country = line.split(':')[0]
pr_value = int(line.split(':')[-1].lstrip().strip())
print(country)

diaosi.update({country: pr_value})

sorted_x = sorted(diaosi.items(), key=operator.itemgetter(1), reverse=True)

with open('fuck.txt', 'a+', encoding="utf-8") as file:
for line in sorted_x:
file.writelines(line[0] + ':' + str(line[1]) + '\n')
对字典key进行排序: sorted_x = sorted(diaosi.items(), key=operator.itemgetter(0))


对keys存在的,对value进行相加
with open('new_domain.txt') as f:
diaosi = {}
for line in f.readlines():
month = int(line.split(':')[1])
pr = line.split(':')[-1].strip()
value = int(diaosi.get(str(month), "0")) + int(pr)
diaosi.update({str(month): value})
print(diaosi)


def has_primary_key():
for row in rows:
if row[1] == 0 and row[9] != 'YES':
return True
return False
非常的简单,但是,如果我们使用any函数的话,代码将会更短。如下所示:
def has_primary_key():
return any(row[1] == 0 and row[9] != 'YES' for row in rows):


刚开始学Python时候帮同事写的一个需求,这几天看看pythonic果然还有更好的写法
# 以长度为统计分别放入不同的列表中
for url in urls:
url_len = str(len(url))
if url_len in url_list:
url_list[url_len].append(url)
else:
url_list[url_len] = [url]

defaultdict实现:
d = defaultdict(list)
for key, value in pairs:
d[key].append(value)


去重复,不改变顺序:

l1 = ['b','c','d','b','c','a','a']
l2 = sorted(set(l1),key=l1.index)
print l2

奇技淫巧倒算不上,有些时候确实是挺有用的!
list_ = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
把list_合并为[1, 2, 3, 4, 5, 6, 7, 8, 9]
[k for i in list_ for k in i]
可以这样:
items = [1, 2, 3, 4, 5]
squared = list(map(lambda x: x**2, items))


  

#coding:utf-8

for line in range(1,255):
with open('1.txt','a') as f:
f.writelines('192.168.128.{}'.format(line)+'\n')


  

  

  

 

  

  
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