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字符串分段组合python

>>> import itertools >>> srcstr = 'abcd' >>> [''.join(x) for x in itertools.permutations(src, 2)] ['ab', 'ac', 'ad', 'ba', 'bc', 'bd', 'ca', 'cb', 'cd', 'da', 'db', 'dc'] >>> [''.join(x) for x in itertools.permutations(src, 3)] ['abc', 'abd', 'acb', 'acd', 'adb', 'adc', 'bac', 'bad', 'bca',

python替换某个文本中的字符串,然后生成新的文本文档,代码如下:import osos.chdir('d:\\') # 跳到d盘if not os.path.exists('test1.txt'): # 看一下这个文件是否存在exit(-1) #不存在就退出lines = open('test1.txt').readlines() #打开文件,读入每

import res = 'abcdefghijkl42819iwlh'm = re.findall(r'[a-eg-za-eg-z]+', s)print(m)运行结果是['abcde', 'ghijkl', 'iwlh']除了f以外的所有字母都匹配到了

如果是字符串直接+就是了.a="a"b="b"c=a+bprint c 如果解决了您的问题请采纳!如果未解决请继续追问

a = "12345678"b = "abcdefg"#生成所2113有组5261合41021653result = [(color, size) for color in a for size in b]import random#随机选取版一个权组合print(random.choice(a),random.choice(b))

import re L=re.findall(r'(//.*?\.room)',a)[willie@localhost ~]$ python3 Python 3.5.2 (default, Dec 7 2016, 23:38:49) [GCC 4.4.7 20120313 (Red Hat 4.4.7-17)] on linux Type "help", "copyright", "credits" or "license" for more information.>>>

Python中提供了3个判断字符串的方法.分别是: 字符串.isdecimal() 字符串.isdigit() 字符串.isnumeric()这三个方法都可以判断字符串是否是由纯阿拉伯数字构成,即0-9组成的数字.这三个方法的区别:字符串.isdecimal() :只能判

去掉两端字符串: strip(), rstrip(),lstrip()123456789101112131415#!/usr/bin/python3 s = ' -----abc123++++ ' # 删除两边空字符print(s.strip()) # 删除左边空字符print(s.rstrip()) # 删除右边空字符print(s.lstrip()) # 删除两边 - + 和空字符print(s.strip().strip('-+'

# This is the basic DFS recursive programming.tab=['a','b','c',..] # wirte the character list heredef helper(depth,ls,lls): if(depth==3): lls.append(list(ls)) return lls.append(list(ls)) for c in tab: ls.append(c) helper(depth+1,ls,lls) ls.pop()lls=[]helper(0,[],lls)print(lls)

>> from itertools import combinations, permutations>> permutations([1, 2, 3], 2)<itertools.permutations at 0x7febfd880fc0> # 可迭代对象 >> list(permutations([1, 2, 3], 2)) #排列[(1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2)] >> list(combinations([1, 2, 3], 2)) #组合[(1, 2), (1, 3), (2, 3)]

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