Question

我有一个字母列表，我想看看这些字母是否出现在状态列表中。如果它们出现，我想知道出现哪个字母以及它出现在哪个位置。我想将它存储在变量中，以便我可以将它与另一个字符串进行比较。以下是我的示例代码：

    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

试试这个：

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

你可以做这样的事情来获得信件的所有索引的列表

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

Answer 3

您可以使用list comprehensions！

列表推导提供了创建列表的简明方法。共同应用程序将创建新列表，其中每个元素都是结果一些操作应用于另一个序列的每个成员或可迭代的，或创建满足a的元素的子序列某种条件。

对于每个状态，对于每个字符，检查它是否存在于您的字母列表中。如果找到，那么您可以在列表中获取索引和字符而不会丢失其顺序！

>>> [(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')])]

如果你打破列表理解，

>>> [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')])]

为了更容易访问元素，您可以使用dict理解！

>>>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')]}

所以当你想要访问某个州时说'佛罗里达'

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

希望它有所帮助！

Answer 4

你可以使用很多方法：）

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

使用后：

print search_letters(list_of_letters)

您将收到如下内容：

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

因此，您将收到类似于此问题说明中描述的内容。当然，如果需要，您可以对特定字典键的列表进行排序。

Python 3：如何在列表中找到char并获取出现次数和位置？

4 个答案: