Create a compress function in Python?

问题

I need to create a function called compress that compresses a string by replacing any repeated letters with a letter and number. My function should return the shortened version of the string. I've been able to count the first character but not any others.

Ex:

>>> compress("ddaaaff")
'd2a3f2'


 def compress(s):
     count=0

     for i in range(0,len(s)):
         if s[i] == s[i-1]:
             count += 1
         c = s.count(s[i])

     return str(s[i]) + str(c)

回答1:

Here is a short python implementation of a compression function:

def compress(string):

    res = ""

    count = 1

    #Add in first character
    res += string[0]

    #Iterate through loop, skipping last one
    for i in range(len(string)-1):
        if(string[i] == string[i+1]):
            count+=1
        else:
            if(count > 1):
                #Ignore if no repeats
                res += str(count)
            res += string[i+1]
            count = 1
    #print last one
    if(count > 1):
        res += str(count)
    return res

Here are a few examples:

>>> compress("ddaaaff")
'd2a3f2'
>>> compress("daaaafffyy")
'da4f3y2'
>>> compress("mississippi")
'mis2is2ip2i'

回答2:

Short version with generators:

from itertools import groupby
def compress(string):
    return ''.join('%s%s' % (char, sum(1 for _ in group)) for char, group in groupby(string)).replace('1', '')

(1) Grouping by chars with groupby(string)

(2) Counting length of group with sum(1 for _ in group) (because no len on group is possible)

(3) Joining into proper format

(4) Removing 1 chars for single items

回答3:

There are several reasons why this doesn't work. You really need to try debugging this yourself first. Put in a few print statements to trace the execution. For instance:

def compress(s):
    count=0

    for i in range(0, len(s)):
        print "Checking character", i, s[i]
        if s[i] == s[i-1]:
            count += 1
        c = s.count(s[i])
        print "Found", s[i], c, "times"

    return str(s[i]) + str(c)

print compress("ddaaaff")

Here's the output:

Checking character 0 d
Found d 2 times
Checking character 1 d
Found d 2 times
Checking character 2 a
Found a 3 times
Checking character 3 a
Found a 3 times
Checking character 4 a
Found a 3 times
Checking character 5 f
Found f 2 times
Checking character 6 f
Found f 2 times
f2

Process finished with exit code 0

(1) You throw away the results of all but the last letter's search. (2) You count all occurrences, not merely the consecutive ones. (3) You cast a string to a string -- redundant.

Try working through this example with pencil and paper. Write down the steps you use, as a human being, to parse the string. Work on translating those to Python.

回答4:

x="mississippi"
res = ""
count = 0
while (len(x) > 0):
    count = 1
    res= ""
    for j in range(1, len(x)):
        if x[0]==x[j]:
            count= count + 1
        else:
            res = res + x[j]
    print(x[0], count, end=" ")
    x=res

回答5:

Just another simplest way to perform this:

def compress(str1):
    output = ''
    initial = str1[0]
    output = output + initial
    count = 1
    for item in str1[1:]:
        if item == initial:
            count = count + 1
        else:
            if count == 1:
                count = ''
            output = output + str(count)
            count = 1
            initial = item
            output = output + item
    print (output)

Which gives the output as required, examples:

>> compress("aaaaaaaccddddeehhyiiiuuo")
a7c2d4e2h2yi3u2o

>> compress("lllhhjuuuirrdtt")
l3h2ju3ir2dt

>> compress("mississippi")
mis2is2ip2i

回答6:

input = "mississippi"
count = 1
for i in range(1, len(input) + 1):
    if i == len(input):
        print(input[i - 1] + str(count), end="")
        break
    else:
        if input[i - 1] == input[i]:
            count += 1
    else:
            print(input[i - 1] + str(count), end="")
            count = 1

Output : m1i1s2i1s2i1p2i1

回答7:

s=input("Enter the string:")
temp={}
result=" "
for x in s:
    if x in temp:
        temp[x]=temp[x]+1
    else:
        temp[x]=1
for key,value in temp.items():
    result+=str(key)+str(value)

print(result)

回答8:

Here is something I wrote.

def stringCompression(str1):
  counter=0
  prevChar = str1[0]
  str2=""
  charChanged = False
  loopCounter = 0

  for char in str1:
      if(char==prevChar):
          counter+=1
          charChanged = False
      else:
          str2 += prevChar + str(counter)
          counter=1
          prevChar = char
          if(loopCounter == len(str1) - 1):
              str2 += prevChar + str(counter)
          charChanged = True
      loopCounter+=1
  if(not charChanged):
      str2+= prevChar + str(counter)

  return str2

Not the best code I guess. But works well.

a -> a1

aaabbbccc -> a3b3c3

回答9:

This is a solution to the problem. But keep in mind that this method only effectively works if there's a lot of repetition, specifically if consecutive characters are repetitive. Otherwise, it will only worsen the situation.

e.g.,
AABCD --> A2B1C1D1
BcDG ---> B1c1D1G1

def compress_string(s):
    result = [""] * len(s)
    visited = None

    index = 0
    count = 1

    for c in s:
        if c == visited:
            count += 1
            result[index] = f"{c}{count}"
        else:
            count = 1
            index += 1
            result[index] = f"{c}{count}"
            visited = c

    return "".join(result)

来源：https://stackoverflow.com/questions/32855812/create-a-compress-function-in-python

标签

python

compression