计算字符串中出现的字符数

Question

I want to find out how often does "reindeer" (in any order) come in a random string and what is the left over string after "reindeer" is removed. I need to preserve order of the left over string

So for example

"erindAeer" -> A (reindeer comes 1 time)

"ierndeBeCrerindAeer" -> ( 2 reindeers, left over is BCA)

I thought of sorting and removing "reindeer", but i need to preserve the order . What's a good way to do this?

Answer 1

Here is the code in Python:

def find_reindeers(s):
    rmap = {}
    for x in "reindeer":
        if x not in rmap:
            rmap[x] = 0
        rmap[x] += 1

    hmap = {key: 0 for key in "reindeer"}
    for x in s:
        if x in "reindeer":
            hmap[x] += 1

    total_occ = min([hmap[x]//rmap[x] for x in "reindeer"])

    left_over = ""
    print(hmap, rmap)
    for x in s:
        if (x in "reindeer" and hmap[x] > total_occ * rmap[x]) or (x not in "reindeer"):
            left_over += x

    return total_occ, left_over

print(find_reindeers("ierndeBeCrerindAeer"))

Output for ierndeBeCrerindAeer :

(2, "BCA")

Answer 2

Here is a rather simple approach using collections.Counter :

from collections import Counter

def purge(pattern, string):
    scount, pcount = Counter(string), Counter(pattern)
    cnt = min(scount[x] // pcount[x] for x in pcount)
    scount.subtract(pattern * cnt)
    return cnt, "".join(scount.subtract(c) or c for c in string if scount[c])

>>> purge("reindeer", "ierndeBeCrerindAeer")
(2, 'BCA')

Answer 3

We can replace those letters after knowing how many times they repeat, and Counter is convenient for counting elements.

from collections import Counter

def leftover(letter_set, string):
    lcount, scount = Counter(letter_set), Counter(string)
    repeat = min(scount[l] // lcount[l] for l in lcount)
    for l in lcount:
        string = string.replace(l, "", lcount[l] * repeat)
    return f"{repeat} {letter_set}, left over is {string}"

print(leftover("reindeer", "ierndeBeCrerindAeer"))
print(leftover("reindeer", "ierndeBeCrerindAeere"))
print(leftover("reindeer", "ierndeBeCrerindAee"))

Output:

2 reindeer, left over is BCA
2 reindeer, left over is BCAe
1 reindeer, left over is BCerindAee

Answer 4

You can do it by using count and replace string function:

import queue
word = "reindeer" 
given_string = "ierndeBeCrerindAeer"
new_string = ""
counter = 0
tmp = ""
letters = queue.Queue()



for i in given_string:
    if not i in word:
        new_string += i
    else:
        letters.put(i)


x = 0
while x < len(word):
    while not letters.empty():
        j = letters.get()
        if j == word[x]:
            tmp += j
            # print(tmp)
            break
        else:
            letters.put(j)
    x = x +1
    if tmp == word:
        counter += 1
        tmp = ""
        x = 0
print(f"The word {word} occurs {counter} times in the string {given_string}.")
print("The left over word is",new_string)

Output will be:

The word reindeer occurs 2 times in the string ierndeBeCrerindAeer.
The left over word is BCA

It's easy to use queue here so that we don't repeat the elements that are already present or found. Hope this answers your question, Thank you!