有一个请求，判断字符串里每个字符出现的次数
python
解体思路：列表的count方法
Python 2.6.6 (r266:84292, Aug 9 2016, 06:11:56)
[GCC 4.4.7 20120313 (Red Hat 4.4.7-17)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> var="dsvsdwefsdagafoijkafdgweodfas"#要判断的字符串
>>> varlist=list(var)#列表化字符串
>>> print varlist
['d', 's', 'v', 's', 'd', 'w', 'e', 'f', 's', 'd', 'a', 'g', 'a', 'f', 'o', 'i', 'j', 'k', 'a', 'f', 'd', 'g', 'w', 'e', 'o', 'd', 'f', 'a', 's']
>>> sinlist=[]#要一个空列表
>>> for letter in varlist:#排除重复字符，把不重复的字符装入新列表。
... if letter not in sinlist:
... sinlist.append(letter)
...
>>> print sinlist
['d', 's', 'v', 'w', 'e', 'f', 'a', 'g', 'o', 'i', 'j', 'k']#这是不重复的列表
>>> for letter in sinlist:#在新列表里循环，用list.count的方法确认每个字符在旧列表出现的次数
... print "letter name is " + letter + " count: " + str(varlist.count(letter))
...
letter name is d count: 5
letter name is s count: 4
letter name is v count: 1
letter name is w count: 2
letter name is e count: 2
letter name is f count: 4
letter name is a count: 4
letter name is g count: 2
letter name is o count: 2
letter name is i count: 1
letter name is j count: 1
letter name is k count: 1
>>> quit()
shell
解题思路：字符串长度，cut和grep
[myname@mypc Desktop]$ var="dsvsdwefsdagafoijkafdgweodfas"#要判断的字符串
[myname@mypc Desktop]$ endnum=`echo -n $var|wc -c`#计算出字符串长度，记住这里用的是echo -n为了排除换行符
[myname@mypc Desktop]$ echo $endnum
29
[myname@mypc Desktop]$ i=1#赋值一个int变量
[myname@mypc Desktop]$ while [ $i -le $endnum ];do echo $var|cut -c $i;i=`expr $i + 1`;done>tempfile#用一个循环把每个字符打印出来，重定向一个临时文件
[myname@mypc Desktop]$ cat tempfile
d
s
v
s
d
w
e
f
s
d
a
g
a
f
o
i
j
k
a
f
d
g
w
e
o
d
f
a
s
[myname@mypc Desktop]$ i=1
[myname@mypc Desktop]$ while [ $i -le $endnum ];do letter=`echo $var|cut -c $i`;count=`grep $letter tempfile|wc -l`;echo "letter is $letter,count is $count";i=`expr $i + 1`;done|sort -u
letter is a,count is 4
letter is d,count is 5
letter is e,count is 2
letter is f,count is 4
letter is g,count is 2
letter is i,count is 1
letter is j,count is 1
letter is k,count is 1
letter is o,count is 2
letter is s,count is 4
letter is v,count is 1
letter is w,count is 2
#再用一个循环去判断每个字符出现的次数
详细解析一下这个循环
while [ $i -le $endnum ]#当i小于或等于字符串长度的时候循环
do
letter=`echo $var|cut -c $i`#取出字符名字
count=`grep $letter tempfile|wc -l`#用grep取出tempfile里字符出现的行数
echo "letter is $letter,count is $count"#打印出结果
i=`expr $i + 1`#i递增
done|sort -u#sort -u是去重，因为里面有重复的结果
perl
perl我不是很懂，查了半天，对于perl我的评价是好用是好用，但是固有写法中符号用的太多，很容易让初学者看起来像天书
解题思路：字符转数组split,替换s/var/var/g，听说还能用hash解，不会中
[myname@mypc Desktop]$ perl -e 'my $var="dsfesfsavcxgsatewqrsadfsaga";#赋值var
> my @varlist=split("",$var);#转化var为数组varlist
> my @grepaf = grep{++$count{$_}<2} @varlist;#这个网上抄来的去重，原理是循环数组varlist，当计数不重复的时候记录，重复不计入数组grepaf
> foreach $a (@grepaf) #在不重复的数组里循环
> { $count=$var=~s/$a/$a/g;#这个意思是说替换变量a为变量a，然后替换的次数就成了count的值。
> print $a," is ",$count,"\n"; }'#打印结果
d is 2
s is 6
f is 3
e is 2
a is 5
v is 1
c is 1
x is 1
g is 2
t is 1
w is 1
q is 1
r is 1
对比下来，python的方法最简单，shell的最多步骤，perl写的最复杂（perl自带的符号太多了，学习这些符号好累）
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