计算python中列表的每个序列中元素的出现

Question

I have a (very big) list like the small example. I want to count the number of D in each sequence in the list and divide by the length of that sequence. (occurrence of D in each sequence).

small example:

l = ['MLSLLLLDLLGLG', 'MEPPQETNRPFSTLDD', 'MVDLSVSPDVPKPAVI', 'XNLMNAIMGSDDDG', 'MDRAPTEQNDDVKLSAE']

do you guys know how to do that?

Answer 1

You can simply use a list comprehension , get the count of D in each sequence and divide by the length of the sequence:

l = ['MLSLLLLDLLGLG', 'MEPPQETNRPFSTLDD', 'MVDLSVSPDVPKPAVI', 'XNLMNAIMGSDDDG', 'MDRAPTEQNDDVKLSAE']

result = [x.count('D')/len(x) for x in l]
print(result)
# [0.07692307692307693, 0.125, 0.125, 0.21428571428571427, 0.17647058823529413]

To handle zero length sequences and avoid ZeroDivisionError , you may use a ternary operator :

result = [(x.count('D')/len(x) if x else 0) for x in l]

Answer 2

You can use list comprehension in order to get the expected result.

I have iterated over each item in the list, for each one of the items in the list I've counted the number of the occurrences of the specified sub string (in this case 'D').

Last, I've divided the number of the occurrences with the length of the item.

l = ['MLSLLLLDLLGLG', 'MEPPQETNRPFSTLDD', 'MVDLSVSPDVPKPAVI', 'XNLMNAIMGSDDDG', 'MDRAPTEQNDDVKLSAE']
output = [float(item.count("D")) / float(len(item)) for item in l]

Answer 3

You're looking for something like:

def fn(x):
    return x.count('D') / len(x)
results = ap(fn, l)