python基础技巧综合训练题2

1，判断一个字符串中的每一个字母是否都在另一个字符串中，可以利用集合的特性来解，集合的元素如果存在，再次更新(update) 是添加不进集合的，那么集合的长度还是跟原来一样，如果添加进去，集合长度就会增加

>>> a = ‘ghost‘
>>> b = ‘hello, can you help me install ghost windows xp system‘
>>> b_set = set( b )
>>> b_set.update( list( a ) )
>>> print len( b_set ) == len( set( b ) )
True
>>> a = ‘abcostg‘
>>> b_set.update( list( a ) )
>>> print len( b_set ) == len( set( b ) )
False
>>> 

2，如果是多个字符呢?

#!/usr/bin/python
#coding:utf-8

#str_list = [ ‘abc‘, ‘ghost‘, ‘hello‘ ]
str_list = [ ‘abc‘, ‘ghost‘, ‘hellox‘ ]
target_str = "abcdefghijklopqrst"
target_str_set = set( target_str )

for val in str_list:
    target_str_set.update( val )

print len( target_str_set ) == len( set( target_str ) )

3，统计出现次数最多的字符

[email protected]:~/python/tmp$ python str3.py
[(‘f‘, 7), (‘s‘, 5), (‘a‘, 4), (‘j‘, 4), (‘k‘, 3), (‘h‘, 2), (‘3‘, 2), (‘1‘, 2), (‘2‘, 2), (‘d‘, 1), (‘l‘, 1), (‘4‘, 1), (‘;‘, 1)]
[email protected]:~/python/tmp$ cat str3.py
#!/usr/bin/python
#coding:utf-8

str = ‘askfjkjasf1234fasdfasfsh;lkjfhjf123‘

l = ( [ ( key, str.count( key ) ) for key in set( str ) ] )
l.sort( key = lambda item : item[1], reverse = True )
print l

[email protected]:~/python/tmp$ 

这里有个lambda表达式, key指定按哪个键排序， item是形参，代表当前的元组，item[1]，那就是取元组中第2项，这里就是字符串的次数，reverse = True,从高到低排序 .

4，统计this模块中, be, is, than,三个单词的出现次数

[email protected]:~/python/tmp$ !p
python statics.py
[(‘be‘, 3), (‘is‘, 10), (‘than‘, 8)]
[email protected]:~/python/tmp$ cat statics.py
#!/usr/bin/python
#coding:utf-8

import os
this_str = os.popen( "python -m this" ).read()
this_str = this_str.replace( ‘\n‘, ‘‘ )
l = this_str.split( ‘ ‘ )

print [ ( x, l.count( x ) ) for x in [‘be‘, ‘is‘, ‘than‘ ] ]
[email protected]:~/python/tmp$ 

os.popen( "python -m this" ).read  读出命令行python -m this  模块的执行结果到一个字符串中

5，用位移运算符，换算b, kb, mb之间的转换关系

[email protected]:~/software$ ls -l sogoupinyin_2.2.0.0102_amd64.deb
-rw-rw-r-- 1 ghostwu ghostwu 22852956 2月   2 14:36 sogoupinyin_2.2.0.0102_amd64.deb
[email protected]:~/software$ ls -lh sogoupinyin_2.2.0.0102_amd64.deb
-rw-rw-r-- 1 ghostwu ghostwu 22M 2月   2 14:36 sogoupinyin_2.2.0.0102_amd64.deb
[email protected]:~/software$ python
Python 2.7.12 (default, Dec  4 2017, 14:50:18)
[GCC 5.4.0 20160609] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> size = 22852956
>>> print "%s kb" % ( size >> 10 )
22317 kb
>>> print "%s MB" % ( size >> 20 )
21 MB
>>> 

6，把列表中的值，连接成字符串

>>> a = [10, 20, 30, 1, 2, 3]
>>> s = str( a )
>>> s
‘[10, 20, 30, 1, 2, 3]‘
>>> type( s )
<type ‘str‘>
>>> s[1:-1]
‘10, 20, 30, 1, 2, 3‘
>>> s.replace( ‘, ‘, ‘‘, s[1:-1] )
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: an integer is required
>>> s[1:-1].replace( ‘, ‘, ‘‘ )
‘102030123‘
>>> 

原文地址：https://www.cnblogs.com/ghostwu/p/8684509.html