如何将多个 string.counts 添加到列表中？ 我正在尝试使用元组打印数字

Question

dna_string = 'ATGCTTCAGAAAGGTCTTACG'

length = len(dna_string)
print("There are %d letters in this DNA string." % length)

print('Now here are the amounts for the letters "A", "C", "T", "G" in order.\n')

combien_a = dna_string.count('A')
combien_c = dna_string.count('C')
combien_t = dna_string.count('T')
combien_g = dna_string.count('G')

print(str(combien_a) + ' ' + str(combien_c) + ' ' + str(combien_g) + ' ' + str(combien_t))

Answer 1

you can try this.

dna_string = 'ATGCTTCAGAAAGGTCTTACG'

print(*[dna_string.count(a) for a in ['A','C','T','G']],sep=" ")

Answer 2

You can add these to a list like:

combien = [dna_string.count(x) for x in ['A','C','T','G']]

Answer 3

collections.Counter is useful for counting numbers.

from collections import Counter

dna_string = 'ATGCTTCAGAAAGGTCTTACG'
# Counter object works like a dictionary with element as key and count as value
combien = Counter(dna_string)
print(f"There are {len(dna_string)} letters in this DNA string.")
# you can convert the Counter to a list of (elem, count) tuples
print(list(combien.items())

Output:

There are 21 letters in this DNA string.
[('A', 6), ('T', 6), ('G', 5), ('C', 4)]

Elements are ordered from higher count to lower count in Counter, if you want another order for the result you may sort like this:

print(sorted(list(combien.items()), key=lambda x: ["A", "C", "T", "G"].index(x[0])))

Output:

[('A', 6), ('C', 4), ('T', 6), ('G', 5)]