如何计算列表中嵌套列表/字典较多的给定字符串的出现？

Question

During a technical phone screen, I was given a list similar to below and asked to count the number of times the string 'a' occurs:

input = ['a', 'b', ['a', 'b', {'a': ['b', 'a', [{'b': ['a', 'b', {'a': 'a'}]}]]}]]

As you can see, in addition to the 'a' being an item in the list, it can also be an item in nested lists, as well as the keys and/or values in nested dictionaries.

This is the code I have so far in Python:

def count_letter_a(arr):

    count = 0

    for item in arr:
        if item == 'a':
            count += 1
        elif isinstance(item, list):
            count_letter_a(item)
        elif isinstance(item, dict):
            for k, v in item.items():
                pass

    return count

I'm stuck on what to do with the handling of dictionary keys/values portion of my function. What do I need to do?

Answer 1

You can just add the 'a' count of keys and values, or simple recursively apply it to the items directly. This requires that you count occurrences in tuples as well (I included sets to complete the built-in collections):

def count_a(obj):
    if isinstance(obj, (list, tuple, set)):
        return sum(map(count_a, obj))
    if isinstance(obj, dict):
        return sum(map(count_a, obj.items()))
    if obj == 'a':
        return 1
    return 0

>>> x = ['a', 'b', ['a', 'b', {'a': ['b', 'a', [{'b': ['a', 'b', {'a': 'a'}]}]]}]]
>>> count_a(x)
7

Answer 2

You're really close!

def count (x):
  if isinstance (x, list):
    return sum (count (y) for y in x)
  elif isinstance (x, dict):
    return sum (count ([ k, v ]) for k,v in x.items ())
  elif x == "a":
    return 1
  else:
    return 0

It works like this

count ('foo')
# 0

count ([ 'a', 'b', 'c' ])
# 1

count (['a', 'b', ['a', 'b', {'a': ['b', 'a', [{'b': ['a', 'b', {'a': 'a'}]}]]}]])
# 7