Python字符串操做

Python 如何判断一个字符串只包含数字字符

Q:如何判断一个字符串只包含数字字符html

A:一种方法是 a.isdigit()。但这种方法对于包含正负号的数字字符串无效,所以更为准确的为:python

    try:
        x = int(aPossibleInt)
        ... do something with x ...
    except ValueError:
        ... do something else ...

这样更准确一些,适用性也更广。但若是你已经确信没有正负号,使用字符串的isdigit()方法则更为方便。git

Python 字符串比较

 

Python 字符串简单比较

简单比较是用内置函数 cmp() 来比较两个字符串:正则表达式

Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>> a = "abc"
>>> b = "abc"
>>> cmp(a, b)
0
>>> c = "def"
>>> cmp(a,c)
-1
>>>

Python字符串比较忽略大小写

 

正则表达式,使用IGNORECASE标志

>
>>> import re
>>> m = re.search('multi', 'A mUltiCased string', re.IGNORECASE)
>>> bool(m)
True

在比较前把2个字符串转换成一样大写

在比较前把2个字符串转换成一样大写,用upper()方法,或小写,lower()
>>> s = 'A mUltiCased string'.lower()
>>> s
'a multicased string'
>>> s.find('multi')
2

python 字符串高级比较

使用python库difflib能够实现两个字符串的比较,找到相同的部分app

Python difflib|SequenceMatcher|Differ|HtmlDiff 使用方法

 

介绍

difflib 是python提供的比较序列(string list)差别的模块。 
实现了三个类: less

  • SequenceMatcher 任意类型序列的比较 (能够比较字符串)
  • Differ 对字符串进行比较
  • HtmlDiff 将比较结果输出为html格式

SequenceMatcher 实例

 

代码:

import difflib
from pprint import pprint
 
a = 'pythonclub.org is wonderful'
b = 'Pythonclub.org also wonderful'
#构造SequenceMatcher类
s = difflib.SequenceMatcher(None, a, b)
 
#获得相同的block
print "s.get_matching_blocks():"
pprint(s.get_matching_blocks())
print 
print "s.get_opcodes():"
for tag, i1, i2, j1, j2 in s.get_opcodes():
    print ("%7s a[%d:%d] (%s) b[%d:%d] (%s)" %  (tag, i1, i2, a[i1:i2], j1, j2, b[j1:j2]))
    #在此实现你的功能

输出为:

s.get_matching_blocks():
[(1, 1, 14), (16, 17, 1), (17, 19, 10), (27, 29, 0)]

s.get_opcodes():
replace a[0:1] (p) b[0:1] (P)
  equal a[1:15] (ythonclub.org ) b[1:15] (ythonclub.org )
replace a[15:16] (i) b[15:17] (al)
  equal a[16:17] (s) b[17:18] (s)
 insert a[17:17] () b[18:19] (o)
  equal a[17:27] ( wonderful) b[19:29] ( wonderful)

SequenceMatcher find_longest_match BUG

import difflib
 
str1 = "Poor Impulse Control: A Good Babysitter Is Hard To Find"
 
str2 = """     A Good Babysitter Is Hard To Find    This is Frederick
by Leo Lionni, the first book I picked for myself.
I was in kindergarten, I believe, which would be either 1968 or 1969.
Frederick has a specific lesson for children about how art is as
important in life as bread, but there's a secondary consideration
I took away: if we pool our talents our lives are immeasurably better.
Curiously, this book is the story of my life, however one interprets
those things. I expect Mickey Rooney to show up any time with a barn
and a plan for a show, though my mom is not making costumes. My sisters
own a toy store with a fantastic selection of imaginative children's books.
I try not to open them because I can't close them and put them back.
My tantrums are setting a bad example for the kids. Anyway, I mention
this because yesterday was Mr. Rogers' 40th anniversary. I appreciate
the peaceful gentleman more as time passes, as I play with finger puppets
in department meetings, as I eye hollow trees for Lady Elaine Fairchild
infestations. Maybe Pete can build me trolley tracks!Labels: To Take
Your Heart Away   """
 
s = difflib.SequenceMatcher(None, str1, str2)
print len(str1), len(str2)
star_a, start_b, length = s.find_longest_match(0, len(str1)-1, 0, len(str2)-1)
print star_a, start_b, length
print str1[star_a:star_a + length]

输出结果为:ide

55 1116
0 1048 1
P

版本为:
Python 2.5.1 (r251:54863, Apr 18 2007, 08:51:08) [MSC v.1310 32 bit (Intel)] on
win32
Type "help", "copyright", "credits" or "license" for more information.
>>>

而最长的应该为 A Good Babysitter Is Hard To Find.函数

解决方法

将 str1 于 str2 交换一下, len(str1) > len(str2). 
则输出结果是想获得的结果。 ui

下面列出了经常使用的python实现的字符串操做this

1.复制字符串

#strcpy(sStr1,sStr2)
sStr1 = 'strcpy'
sStr2 = sStr1
sStr1 = 'strcpy2'
print sStr2

2.链接字符串

#strcat(sStr1,sStr2)
sStr1 = 'strcat'
sStr2 = 'append'
sStr1 += sStr2
print sStr1

3.查找字符

#strchr(sStr1,sStr2)
sStr1 = 'strchr'
sStr2 = 'r'
nPos = sStr1.index(sStr2)
print nPos

4.比较字符串

#strcmp(sStr1,sStr2)
sStr1 = 'strchr'
sStr2 = 'strch'
print cmp(sStr1,sStr2)

5.扫描字符串是否包含指定的字符

#strspn(sStr1,sStr2)
sStr1 = '12345678'
sStr2 = '456'
#sStr1 and chars both in sStr1 and sStr2
print len(sStr1 and sStr2)

6.字符串长度

#strlen(sStr1)
sStr1 = 'strlen'
print len(sStr1)

7.将字符串中的小写字符转换为大写字符

#strlwr(sStr1)
sStr1 = 'JCstrlwr'
sStr1 = sStr1.upper()
print sStr1

8.追加指定长度的字符串

#strncat(sStr1,sStr2,n)
sStr1 = '12345'
sStr2 = 'abcdef'
n = 3
sStr1 += sStr2[0:n]
print sStr1

9.字符串指定长度比较

#strncmp(sStr1,sStr2,n)
sStr1 = '12345'
sStr2 = '123bc'
n = 3
print cmp(sStr1[0:n],sStr2[0:n])

10.复制指定长度的字符

#strncpy(sStr1,sStr2,n)
sStr1 = ''
sStr2 = '12345'
n = 3
sStr1 = sStr2[0:n]
print sStr1

11.字符串比较,不区分大小写

#stricmp(sStr1,sStr2)
sStr1 = 'abcefg'
sStr2 = 'ABCEFG'
print cmp(sStr1.upper(),sStr2.upper())

12.将字符串前n个字符替换为指定的字符

#strnset(sStr1,ch,n)
sStr1 = '12345'
ch = 'r'
n = 3
sStr1 = n * ch + sStr1[3:]
print sStr1

13.扫描字符串

#strpbrk(sStr1,sStr2)
sStr1 = 'cekjgdklab'
sStr2 = 'gka'
nPos = -1
for c in sStr1:
    if c in sStr2:
        nPos = sStr1.index(c)
        break
print nPos

14.翻转字符串

#strrev(sStr1)
sStr1 = 'abcdefg'
sStr1 = sStr1[::-1]
print sStr1

15.查找字符串

python strstr

#strstr(sStr1,sStr2)
sStr1 = 'abcdefg'
sStr2 = 'cde'
print sStr1.find(sStr2)

16.分割字符串

#strtok(sStr1,sStr2)
sStr1 = 'ab,cde,fgh,ijk'
sStr2 = ','
sStr1 = sStr1[sStr1.find(sStr2) + 1:]
print sStr1