如何计算 python 中字母的出现次数？

Question

I have written a custom python code to count the occurrence of an alphabet after a certain alphabet. For example, in my case, I have a string "ABACCD". Now I want to find that starting with alphabet "A", in my string, how many times it is followed by "B", how many times it is followed by "C" etc. Similarly for "B" as well, how many time it is followed by "A", how many time it is followed by "B", how many times it is followed by "C". So the output will be something like: If my alphabet is "A", it is being followed by "A" zero times, it is being followed by "B" 1 time, followed by "C" 1 time and followed by "D" 0 times. The code that I have written just return me a list of 1s. Here is my code:

s = "ABACCD"
a = []

for i in s:
    if(i=="A"):
        for j in s:
            a_count = 0
            if(j=="A"):
                a_count+=1
                a.append(a_count)
print(a)

Any help would be much appreciated. Thanks

Answer 1

It's better of use a dictionary than a list to keep track of all the pairs of letter occurences

1. Basic Python using dictionary

cnts = {}
for a, b in zip(s, s[1:]):
    cnts[(a,b)] = cnts.get((a, b), 0) + 1

print(cnts) # {('A', 'B'): 1, ('B', 'A'): 1, ('A', 'C'): 1, ('C', 'C'): 1, ('C', 'D'): 1}

2. Basic Python without zip

cnts = {}
i = 0
while i < len(s) - 1:
    a = s[i]
    b = s[i+1]
    cnts[(a,b)] = cnts.get((a, b), 0) + 1
    i += 1    
print(cnts)
#{('A', 'B'): 1, ('B', 'A'): 1, ('A', 'C'): 1, ('C', 'C'): 1, ('C', 'D'): 1}

3. Solution using collections.Counter

from collections import Counter
cnts = Counter(zip(s, s[1:]))
print(cnts) 
#Counter({('A', 'B'): 1,
          ('B', 'A'): 1,
          ('A', 'C'): 1,
          ('C', 'C'): 1,
          ('C', 'D'): 1})