如何创建内置函数计数，而无需使用它。 蟒蛇

Question

I would like to get the exact same result as this code:

def histogram(s):
    d = {}

    for w in s: 
        d[w] = s.count(w)

    for k in sorted(d):
        print (k + ': ' + str(d[k]))

But without using any built-in functions. I only want to use len() and range() and chr() and ord().

The result of this program when typing in Mississippi is:

M: 1
i: 4
p: 2
s: 4

To clear it all upp!

Write a function histogram(s) that takes a string as a parameter and returns a list with a histogram over the numbers of characters.

And a function histprint should be implemented, that takes a list as the function histogram(s) have returned and writes a table on the screen, containing all the characters that was in the string. The result should look like:

>>> h=histogram('Mississippi') 
>>> histprint(h) 
M: 1 
i: 4 
p: 2 
s: 4

NO built-in functions!

Answer 1

count and Counter are the ways to go but the following should work fine too.

my = "stackoverflow"
d={}
for l in my:
        d[l] = d.get(l,0) + 1 
print d

{'a': 1, 'c': 1, 'e': 1, 'f': 1, 'k': 1, 'l': 1, 'o': 2, 's': 1, 'r': 1, 't': 1, 'w': 1, 'v': 1}

Answer 2

If you would like to avoid any builtins functions as well as methods such as get. You can use try catch which is same as what @d-coder answered.

Basically you could do:

def histogram(word):
    counter = {}
    for char in word:
        try:
            counter[char] = counter[char] + 1
        except KeyError:
            counter[char] = 1
    return counter

def histprint(h):
    for k in h:
        print k, h[k]

This answer is not different from above and is only removing get with try``except .

Also Counter is not a builtin function but a datatype available for particularly this kind of purpose AFAIK.

Answer 3

You can do that by creating a dictionary of letters, and counting the frequency of letters in the word.

• First, create a dictionary of letters, initialized by zero.

 import string d= dict.fromkeys(string.ascii_letters, 0) #d= dict(zip(string.ascii_letters, [0]*52)) # string.ascii_letters: should give you a list of alphabets (UPPER case,lower case). 

Note: Check the comments

Output: 

 {'A': 0, 'C': 0, 'B': 0, 'E': 0, 'D': 0, 'G': 0, 'F': 0, 'I': 0, 'H': 0, 'K': 0, 'J': 0, 'M': 0, >'L': 0, 'O': 0, 'N': 0, 'Q': 0, 'P': 

0, 'S': 0, 'R': 0, 'U': 0, 'T': 0, 'W': 0, 'V': 0, 'Y': 0, 'X': 0, 'Z': 0, 'a': 0, 'c': 0, 'b': 0, 'e': 0, 'd': 0, 'g': 0, 'f': 0, 'i': 0, 'h': 0, 'k': 0, 'j': 0, 'm': 0, 'l': 0, 'o': 0, 'n': 0, 'q': 0, 'p': 0, 's': 0, 'r': 0, 'u': 0, 't': 0, 'w': 0, 'v': 0, 'y': 0, 'x': 0, 'z': 0}

• Then, go through every character and add +1 whenever you find it.

 w= "Mississippi" for x in list(w): d[x]+=1 l= {} for (key, value) in d.iteritems(): if value>0: print key:value 

Output:

 M : 1 i : 4 p : 2 s : 4 

So your function would be:

def histprint(w):
  d= dict.fromkeys(string.ascii_letters, 0)

  for x in list(w):
    d[x]+=1

  l= {}
  for (key, value) in d.iteritems():
    if value>0:
      print key,':',value


histprint("Mississippi")

If you are counting the characters, and you don't distinguish between uppercase and lower case (ie: You assume that M and m refers to same character, and not to Upper 'M' or lower 'm')

Your algorithm would look like:

import string

def histprint(w):
  d= dict.fromkeys(string.ascii_lowercase, 0)

  for x in list(w.lower()):
    d[x]+=1

  l= {}
  for (key, value) in d.iteritems():
    if value>0:
      print key,':',value


histprint("Mississippi")

Answer 4

def histogram(s):
    d = {}    
    for i in s:
        try:
            d[i] += 1
        except KeyError:
            d[i] = 1
    # Do you want to implement sorting also without inbuilt method ?
    # Not added sorting here ! `sorted(d)` will give the sorted keys.
    for k, v in d.iteritems():
        print "{} : {}".format(k, v)