Access next element in a for loop in python

You've missed,

1. Accessing the proper index to find the next element.
2. Using == operator instead of in property to check the value's presence in the list.

Solution:

bar = ['A', 'C', 'D', 'F', 'H']
foo = [['A', '1'], ['C', '5', 'D', '9', 'E'], ['F'], ['G', 'H']]

output_dict = {}

for element in foo:
    for foo_element in element:
        if foo_element in bar:
            try:
                output_dict[foo_element] = element[element.index(foo_element) + 1]
            except:
                output_dict[foo_element] = '-'
print(output_dict)

Output:

{'D': '9', 'H': '-', 'F': '-', 'C': '5', 'A': '1'}

I would normalize the data, i.e. flatten the foo and convert it to a dict of all pairs {item: next_item}.

Then just get the data from this dict.

foo = [['A', '1'], ['C', '5', 'D', '9', 'E'], ['F'], ['G', 'H']]
bar = ['A', 'C', 'D', 'F', 'H']

pdict = {k:v for s in foo for k, v in zip(s, s[1:])}
# print(pdict) # uncomment to check
pay = {k: pdict.get(k, '-') for k in bar}

print(pay) # {'A': '1', 'C': '5', 'D': '9', 'F': '-', 'H': '-'}

As @kuro mentioned, you can use enumerate function or else you can simply iterate over the the indices as below.

foo = [['A', '1'], ['C', '5', 'D', '9', 'E'], ['F'], ['G', 'H']]
bar = ['A', 'C', 'D', 'F', 'H']
pay = {}

for i in bar:
    for j in foo:
        for k in range(len(j)):
            if i == j[k]:
                try:
                    pay[i] = j[k + 1]
                except IndexError:
                    pay[i] = '-'

print(pay)

Output:

{'A': '1', 'C': '5', 'D': '9', 'F': '-', 'H': '-'}

list_of_lists = foo
flattened = [val for sublist in list_of_lists for val in sublist] + ['-']

pay = {}

for bar_val in bar:
    pos = flattened.index(bar_val)
    try:
        pay[bar_val] = int(flattened[pos+1])
    except ValueError as e:
        pay[bar_val] = '-'

print(pay)

Output:

{'A': 1, 'C': 5, 'D': 9, 'F': '-', 'H': '-'}

You can try to do it this way as well:

foo = [['A', '1'], ['C', '5', 'D', '9', 'E'], ['F'], ['G', 'H']]
bar = ['A', 'C', 'D', 'F', 'H']
pay = dict()

# loop through bar
for i in range(len(bar)):
  # loop through foo
  for j in range(len(foo)):
    # loop through elements of foo
    for k in range(len(foo[j])):
      # check if data in foo matches element in bar
      if bar[i] == foo[j][k]:
        # check if data in foo is the last element in the sub-list
        if (len(foo[j]) > k+1):
          pay[bar[i]] = foo[j][k+1]
        else:
          pay[bar[i]] = '-'
      else:
        pass

# iterate through dict and print its contents
for k, v in pay.items():
    print(k, v)

def get_next(elt, sublist):
    
    try:
        return sublist[sublist.index(elt) + 1]
    except:
        return '-'

final_dict = { elt: get_next(elt, sub)  for sub in foo for elt in sub if elt in bar}

print(final_dict) # {'A': '1', 'C': '5', 'D': '9', 'F': '-', 'H': '-'}

Just another solution using generators:

foo = [['A', '1'], ['C', '5', 'D', '9', 'E'], ['F'], ['G', 'H']]

from itertools import chain

def get_pairs(it):
    stack = None
    while True:
        try:
            x = stack if stack else next(it)
            try:
                y = next(it)
                if y.isdigit():
                    yield x, y
                    stack = None
                else:
                    yield x, "-"
                    stack = y
            except StopIteration:
                yield x, "-"
                break
        except StopIteration:
            break

result = dict([item for item in get_pairs(chain.from_iterable(foo))])
print(result)

Which yields

{'A': '1', 'C': '5', 'D': '9', 'E': '-', 'F': '-', 'G': '-', 'H': '-'}