获取两个列表相交的项目索引的Python方法

Question

Say I have two list: one is a string -- 'example' and another is the alphabet. I'd like to find a more pythonic way where every position in the alphabet list each letter of the string list 'example' intersects and put these indices in a list. IE

• e : 4
• x : 23
• a : 0
• m : 12

etc...

So far I have:

import string
alphabet = list(string.ascii_lowercase)
key = list('example')

def convert(string, alphabet):
    table_l = []
    for char in string:
        for letter in alphabet:
            if letter == char:
                table_l.append(alphabet.index(letter))
    return table_l

convert(key, alphabet)

I've tried using set intersection, but the string 'key' can contain more than 1 of each letter, and I'm looking for indices, not which letters match.

So far, the best I've tried is:

for x in key:
    listed.append(set(alphabet).intersection(x))

I've no clue how to append the keys of alphabet where the value intersects with each letter of key.

Thanks

Answer 1

You want a mapping from letters to numbers, so use a mapping data-structure, eg a dict :

>>> alphamap = dict(zip(alphabet, range(len(alphabet)))
>>> alphamap
{'h': 7, 'e': 4, 'g': 6, 'n': 13, 'm': 12, 's': 18, 'x': 23, 'r': 17, 'o': 14, 'f': 5, 'a': 0, 'v': 21, 't': 19, 'd': 3, 'j': 9, 'l': 11, 'b': 1, 'u': 20, 'y': 24, 'q': 16, 'k': 10, 'c': 2, 'w': 22, 'p': 15, 'i': 8, 'z': 25}
>>> def convert(string, map_):
...     return  [map_[c] for c in string]
...
>>> convert('example', alphamap)
[4, 23, 0, 12, 15, 11, 4]

Note, your original approach could be simplified to:

>>> list(map(alphabet.index, 'example'))
[4, 23, 0, 12, 15, 11, 4]

However, using alphabet.index is less efficient than using a mapping (since it has to do a linear search each time rather than a constant-time hash).

Also, note I've iterated over strings directly, no need to put them into a list, strings are sequences just like list objects. They can be iterated over, sliced, etc. However, they are immutable.

Finally, the above approach will fail if there isn't a corresponding value, ie a special, non-alphabetic character.

>>> convert("example!", alphamap)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in convert
  File "<stdin>", line 2, in <listcomp>
KeyError: '!'

This may or may not be desirable. Alternatively, you can approach this by using .get with a default-value, eg:

>>> def convert(string, map_, default=-1):
...     return  [map_.get(c, default) for c in string]
...
>>> convert("example!", alphamap)
[4, 23, 0, 12, 15, 11, 4, -1]

Answer 2

If it's all ascii, something like below should work - convert letter to numeric representation, then subtract 97 as that's 'a' in ascii

a = ord(‘a’)
[ord(c)-a for c in ‘example’.lower()]

Answer 3

Somehow in the same spirit as Guy, what about counting in base 36 (and following DyZ's and mhawke's advices),

>>> a = int('a', 36)
>>> [int(c, 36) - a for c in 'example']
[4, 23, 0, 12, 15, 11, 4]

---

Note that this method is case insensitive, and works if it's all ascii (which appears to be the case since you play with string.ascii_lowercase ).

Answer 4

Use sets.

overlapKeys = set(alphabet) & set(key)
listOfIndices = [alphabet.index(key) for key in overlapKeys]

Also,

key = list('example')

is unneccessary. Strings are lists of characters. Use

key = 'example'

Answer 5

Your example seems a little off... wouldn't x be 23, m 12, etc?

>>> s = 'example'
>>> [(c, string.ascii_lowercase.index(c)) for c in s]    # as a list of tuples
[('e', 4), ('x', 23), ('a', 0), ('m', 12), ('p', 15), ('l', 11), ('e', 4)]

This would be a little inefficient for longer strings because the use of index() effectively makes this an O(n**2) solution.

A better way is to use a lookup dictionary to convert from a character to its index. Because a dict lookup is O(1) the resulting solution will be O(n), which is much better.

# create a dict that maps characters to indices
indices = {c: index for index, c in enumerate(string.ascii_lowercase)}
# perform the conversion
>>> s = 'example'
>>> [(c, indices.get(c, -1)) for c in s]
[('e', 4), ('x', 23), ('a', 0), ('m', 12), ('p', 15), ('l', 11), ('e', 4)]

If you wanted just the indices:

>>> [indices.get(c, -1) for c in s]
[4, 23, 0, 12, 15, 11, 4]