在python中计算列表中的字符串，然后过滤和匹配

Question

我有一个单词列表，使用 python3 我计算每个单词组合之间的字母差异（使用clever diff_summing algorithm from this site）：

import itertools

def diff_letters(a,b):
    return sum ( a[i] != b[i] for i in range(len(a)) )

w = ['AAHS','AALS','DAHS','XYZA']

for x,y in itertools.combinations(w,2):
    if diff_letters(x,y) == 1:
        print(x,y)

打印出来：

AAHS AALS
AAHS DAHS

我的问题：我如何计算和字符串“DAHS”和“AALS”只有一个合作伙伴，而“AAHS”有两个合作伙伴？我将过滤每个target_string 有正好一个 near_matching_word 的方向组合，所以我的最终数据（作为 JSON）看起来像这样：

[
 {
   "target_word": "DAHS",
   "near_matching_word": "AAHS"
 },
 {
   "target_word": "AALS",
   "near_matching_word": "AAHS"
 }
]

（注意 AAHS 没有显示为 target_word）

我有一个使用functools.reduce的版本

import itertools
import functools
import operator

def diff_letters(a,b):
    return sum ( a[i] != b[i] for i in range(len(a)) )

w = ['AAHS','AALS','DAHS','XYZA']

pairs = []
for x,y in itertools.combinations(w,2):
    if diff_letters(x,y) == 1:
        #print(x,y)
        pairs.append((x,y))

full_list = functools.reduce(operator.add, pairs)
for x in full_list:
    if full_list.count(x) == 1:
        print (x)

打印出来的

AALS
DAHS

但是我必须回到我的大名单pairs 才能找到near_matching_word。当然，在我的最终版本中，pairs 列表会更大，target_word 可以是元组 (x,y) 中的第一项或第二项。

Answer 1

即使找到多个对，其他答案也会保留所有对。由于不需要它们，这似乎浪费了内存。这个答案只为每个字符串保留最多一对。

import collections
import itertools

def diff_letters(a,b):
    return sum ( a[i] != b[i] for i in range(len(a)) )

w = ['AAHS','AALS','DAHS','XYZA']

# Marker for pairs that have not been found yet.
NOT_FOUND = object()

# Collection of found pairs x => y. Each item is in one of three states:
# - y is NOT_FOUND if x has not been seen yet
# - y is a string if it is the only accepted pair for x
# - y is None if there is more than one accepted pair for x
pairs = collections.defaultdict(lambda: NOT_FOUND)

for x,y in itertools.combinations(w,2):
    if diff_letters(x,y) == 1:
        if pairs[x] is NOT_FOUND:
            pairs[x] = y
        else:
            pairs[x] = None
        if pairs[y] is NOT_FOUND:
            pairs[y] = x
        else:
            pairs[y] = None

# Remove None's and change into normal dict.
pairs = {x: y for x, y in pairs.items() if y}

for x, y in pairs.items():
    print("Target = {}, Only near matching word = {}".format(x, y))

输出：

Target = AALS, Only near matching word = AAHS
Target = DAHS, Only near matching word = AAHS

Answer 2

您可以使用字典而不是对列表：

pairs = {}
for x, y in itertools.combinations(w, 2):
    if diff_letters(x, y) == 1:
        pairs.setdefault(x, []).append(y)
        pairs.setdefault(y, []).append(x)

result = [{ "target_word": key, "near_matching_word": head, } for key, (head, *tail) in pairs.items() if not tail]

print(result)

输出

[{'target_word': 'AALS', 'near_matching_word': 'AAHS'}, {'target_word': 'DAHS', 'near_matching_word': 'AAHS'}]

在pairs 字典中，键是target_words，值是near_matching_words。然后使用列表推导过滤掉超过 1 个near_matching_word 的那些。

Answer 3

import itertools
import functools
import operator


def diff_letters(a, b):
    return sum(a[i] != b[i] for i in range(len(a)))


w = ['AAHS', 'AALS', 'DAHS', 'XYZA']

pairs = []
for x, y in itertools.combinations(w, 2):
    if diff_letters(x, y) == 1:
        pairs.append((x, y))

full_list = functools.reduce(operator.add, pairs)

result = []

for x in set(full_list):
    if full_list.count(x) == 1:
        pair = next((i for i in pairs if x in i))
        match = [i for i in pair if i != x][0]
        result.append({
            "target_word": x,
            "near_matching_word": match
        })

print(result)

输出：

[{'target_word': 'DAHS', 'near_matching_word': 'AAHS'}, {'target_word': 'AALS', 'near_matching_word': 'AAHS'}]