方案一 统计字符串中的字母频数

import collections

import re

d = collections.defaultdict(int)

S = "testTypecopyri4g3HACBCtcFor the ginpraise.ll be dog ssditsl-+-___*&^%icense()ation."

#convert to lower case

s = S.lower()

#match non-letter chars

rule = re.compile(r"[^a-z]")

#delete non-letter chars

s_result = rule.sub("", s)

#calculate the couter

for c in s_result:

    d[c] += 1

print d

#rank according to the letter

print sorted(d.iteritems(), key = lambda d:d[0])

#rank according to the counter

print sorted(d.iteritems(), key = lambda d:d[1])

方案二 统计字符串中的字母频率

import collections

import re

sum = 0

d = collections.defaultdict(float)

S = "testTypecopyri4g3HACBCtcFor the ginpraise.ll be dog ssditsl-+-___*&^%icense()ation."

#convert to lower case

s = S.lower()

#match non-letter chars

rule = re.compile(r"[^a-z]")

#delete non-letter chars

s_result = rule.sub("", s)

#calculate the couter

for c in s_result:

   d[c] += 1

   #calculate the total number of letters

   sum += 1

print "The total number of letters is: %d"%sum

#calculate the frequency

for c in d:

   d[c] = d[c]/sum

   d[c] = round(d[c], 10)

print d

#rank according to the letter

print sorted(d.iteritems(), key = lambda d:d[0])

print sorted(d.iteritems(), key = lambda d:d[0], reverse = True)

#rank according to the counter

print sorted(d.iteritems(), key = lambda d:d[1])

print sorted(d.iteritems(), key = lambda d:d[1], reverse = True)

方案三 统计文本文件中的字母频数

import collections

import re

d = collections.defaultdict(int)

#open the txt file

myfile = open('test.txt')

for line in myfile:

   #delete the "n" at the end of the line

   line = line.rstrip('n')

   #convert to lower case

   line = line.lower()

   #match non-letter chars

   rule = re.compile(r"[^a-z]")

   #delete non-letter chars

   line_result = rule.sub("", line)

   #calculate the couter

   for c in line_result:

       d[c] += 1

myfile.close()

print d

#rank according to the letter

print sorted(d.iteritems(), key = lambda d:d[0])

#rank according to the counter

print sorted(d.iteritems(), key = lambda d:d[1])

方案四 统计文本文件中的字母频率

import collections

import re

sum = 0

d = collections.defaultdict(float)

#open the txt file

myfile = open('test.txt')

for line in myfile:

   #delete the "n" at the end of the line

   line = line.rstrip('n')

   #convert to lower case

   line = line.lower()

   #match non-letter chars

   rule = re.compile(r"[^a-z]")

   #delete non-letter chars

   line_result = rule.sub("", line)

   #calculate the couter

   for letter_x in line_result:

       d[letter_x] += 1

       sum += 1

myfile.close()

#calculate the total number of letters

print "The total number of letters is: %d"%sum

#calculate the frequency

for c in d:

  d[c] = d[c]/sum

  d[c] = round(d[c], 10)

print d

#rank according to the letter

print sorted(d.iteritems(), key = lambda d:d[0])

print sorted(d.iteritems(), key = lambda d:d[0], reverse = True)

#rank according to the counter

print sorted(d.iteritems(), key = lambda d:d[1])

print sorted(d.iteritems(), key = lambda d:d[1], reverse = True)