Question

我需要制作一个打印出文字中字母频率的程序文件并将该频率与python中的另一个频率进行比较。

到目前为止，我能够打印出一封信的次数，但是我得到的百分比频率是错误的。我认为这是因为我只需要计算我的程序删除所有空格和其他空格中文件中的字母数字符。

def addLetter (x):
    result = ord(x) - ord(a)
    return result


#start of the main program
#prompt user for a file

while True:
    speech = raw_input("Enter file name:")

    wholeFile = open(speech, 'r+').read()
    lowlet = wholeFile.lower()
    letters= list(lowlet)
    alpha = list('abcdefghijklmnopqrstuvwxyz')
    n = len(letters)
    f = float(n)
    occurrences = {}
    d = {}


    #number of letters
    for x in alpha:
        occurrences[x] = letters.count(x)
        d[x] =(occurrences[x])/f
    for x in occurrences:
        print x, occurrences[x], d[x]

这是输出

Enter file name:dems.txt
a 993 0.0687863674148
c 350 0.0242449431976
b 174 0.0120532003325
e 1406 0.0973954003879
d 430 0.0297866444999
g 219 0.015170407315
f 212 0.0146855084511
i 754 0.0522305347742
h 594 0.0411471321696
k 81 0.00561097256858
j 12 0.000831255195345
m 273 0.0189110556941
l 442 0.0306178996952
o 885 0.0613050706567
n 810 0.0561097256858
q 9 0.000623441396509
p 215 0.0148933222499
s 672 0.0465502909393
r 637 0.0441257966196
u 305 0.021127736215
t 1175 0.0813937378775
w 334 0.0231366029371
v 104 0.00720421169299
y 212 0.0146855084511
x 13 0.000900526461624
z 6 0.000415627597672
Enter file name:

程序会在列中打印，但我不确定如何在此处显示。

“a”的频率应为.0878

Answer 1

您可以使用translator recipe删除alpha以外的所有字符。由于这样做会导致letters只包含alpha中的字符，n现在是正确的分母。

然后，您可以使用collections.defaultdict(int)来计算字母的出现次数：

import collections
import string

def translator(frm='', to='', delete='', keep=None):
    # Python Cookbook Recipe 1.9
    # Chris Perkins, Raymond Hettinger
    if len(to) == 1: to = to * len(frm)
    trans = string.maketrans(frm, to)
    if keep is not None:
        allchars = string.maketrans('', '')
        # delete is expanded to delete everything except
        # what is mentioned in set(keep)-set(delete)
        delete = allchars.translate(allchars, keep.translate(allchars, delete))
    def translate(s):
        return s.translate(trans, delete)
    return translate

alpha = 'abcdefghijklmnopqrstuvwxyz'
keep_alpha=translator(keep=alpha)

while True:
    speech = raw_input("Enter file name:")
    wholeFile = open(speech, 'r+').read()
    lowlet = wholeFile.lower()
    letters = keep_alpha(lowlet)
    n = len(letters)
    occurrences = collections.defaultdict(int)    
    for x in letters:
        occurrences[x]+=1
    for x in occurrences:
        print x, occurrences[x], occurrences[x]/float(n)

Answer 2

import collections
import re
from __future__ import division

file1 = re.subn(r"\W", "", open("file1.txt", "r").read())[0].lower()
counter1 = collections.Counter(file1)
for k, v in counter1.iteritems():
   counter1[k] = v / len(file1)

file2 = re.subn(r"\W", "", open("file2.txt", "r").read())[0].lower()
counter2 = collections.Counter(file2)
for k, v in counter2.iteritems():
   counter2[k] = v / len(file2)

注意：需要Python 2.7。

Answer 3

我认为这是一种非常直接的方式：

while True:
    speech = raw_input("Enter file name:")

    wholeFile = open(speech, 'r+').read()
    lowlet = wholeFile.lower()

    alphas = 'abcdefghijklmnopqrstuvwxyz'

    # lets set default values first
    occurrences = {letter : 0 for letter in alphas }
    # occurrences = dict(zip(alphas, [0]*len(alphas))) # for python<=2.6

    # total number of valid letters
    total = 0

    # iter everything in the text
    for letter in lowlet:
        # if it is a valid letter then it is in occurrences
        if letter in occurrences:
            # update counts
            total += 1
            occurrences[letter] += 1

    # now print the results:
    for letter, count in occurrences.iteritems():
        print letter, (1.0*count/total)

如您所知，在计算频率之前，您需要文本中的有效字母总数。要么在处理之前过滤文本，要么将过滤与处理结合起来，这就是我在这里所做的。

python中的字母频率

3 个答案: