Python 3：如何在列表中找到一个字符并获取出现次数和位置？

Question

I have a list of letters, I want to see if these letters appear in a list of states. If they do appear, I want to know which letter appears and in which position it appears. I want to store it in a variable so I can then compare it to another string. Below is my example code:

    letters = ['a','b','c','d','e']
    states = ['minnesota','new york','florida']
    found_states = [] 

    for letter in letters:
        for state in states:
            if letter in state:
            found_states.append(state)
            #Here instead of appending to a list
            #I want to find the position of each letter
            #without losing the letter itself
            found_letter = {'e':4,'a':8} #desired result for minnesota
            #i want to use found_letter variable to perform additional
            #if statements
    print(found_states)

Answer 1

Try this:

found_letter = {}
i = 0
for letter in letters:
    i=i+1
    for state in states:
        if letter in state:
            found_states.append(state)
            if letter in found_letter:
                found_letter[letter].append(i)
            else:
                found_letter[letter] = []
                found_letter[letter].append(i)
print(found_states)

Answer 2

You could do something like this to get a list of all indices of the letter

state = 'ohio'
letter = 'o'
occurences = [i for i, c in enumerate(state) if c == letter]

Answer 3

You can use list comprehensions !

List comprehensions provide a concise way to create lists. Common applications are to make new lists where each element is the result of some operations applied to each member of another sequence or iterable, or to create a subsequence of those elements that satisfy a certain condition.

For every s in states, for every character, check if its present in your letters list. If found, then you can get the index and the character in a list without losing its order!

>>> [(s,[(index,c) for index,c in enumerate(s) if c in letters]) for s in states]
[('minnesota', [(4, 'e'), (8, 'a')]), ('new york', [(1, 'e')]), ('florida', [(5, 'd'), (6, 'a')])]

If you break down the list comprehension,

>>> [s for s in states] 
['minnesota', 'new york', 'florida']

>>> [[c for index,c in enumerate(s)] for s in states] 
[['m', 'i', 'n', 'n', 'e', 's', 'o', 't', 'a'], ['n', 'e', 'w', ' ', 'y', 'o', 'r', 'k'], ['f', 'l', 'o', 'r', 'i', 'd', 'a']]

>>> [[(index,c) for index,c in enumerate(s)] for s in states]
[[(0, 'm'), (1, 'i'), (2, 'n'), (3, 'n'), (4, 'e'), (5, 's'), (6, 'o'), (7, 't'), (8, 'a')], [(0, 'n'), (1, 'e'), (2, 'w'), (3, ' '), (4, 'y'), (5, 'o'), (6, 'r'), (7, 'k')], [(0, 'f'), (1, 'l'), (2, 'o'), (3, 'r'), (4, 'i'), (5, 'd'), (6, 'a')]]

>>> [(s,[(index,c) for index,c in enumerate(s)]) for s in states]
[('minnesota', [(0, 'm'), (1, 'i'), (2, 'n'), (3, 'n'), (4, 'e'), (5, 's'), (6, 'o'), (7, 't'), (8, 'a')]), ('new york', [(0, 'n'), (1, 'e'), (2, 'w'), (3, ' '), (4, 'y'), (5, 'o'), (6, 'r'), (7, 'k')]), ('florida', [(0, 'f'), (1, 'l'), (2, 'o'), (3, 'r'), (4, 'i'), (5, 'd'), (6, 'a')])]

To make access of elements much easier, you can use dict comprehension!

>>>res =  {s:[(index,c) for index,c in enumerate(s) if c in letters] for s in states}
>>> print res
{'minnesota': [(4, 'e'), (8, 'a')], 'new york':[(1, 'e')], 'florida':[(5, 'd'), (6, 'a')]}

So when you wan to access for one state say 'florida'

>>> print res['florida']
[(5, 'd'), (6, 'a')]

Hope it helps!

Answer 4

You can use a lot of approaches :)

list_of_letters = ['a', 'b', 'c', 'd', 'e', 'n']
states = ['minnesota', 'new york', '

    def search_letters(letters):
        found_states = {}
        for letter in letters:
            for state in states:
                for i, char in enumerate(state):
                    if char in letter:
                        found_states.setdefault(state, []).append(letter + ":" + str(i + 1))
        return found_states

After you use:

print search_letters(list_of_letters)

You will receive something like:

{'new york': ['e:2', 'n:1'], 'florida': ['a:7', 'd:6'], 'minnesota': ['a:9', 'e:5', 'n:3', 'n:4']}

So you will receive something like you described in the description of this issue. Of course you can sort the lists of a particular dictionary key, if you want.