[英]Python - For Loop of Itertools Permutations, How to not Create Endless Elif Statments?
我是python /編程新手,已經使用了幾個月。 希望這段代碼對於SO來說不會太大或太高,但是我無法弄清楚在沒有完整上下文的情況下如何問這個問題。 因此,這里去:
import re
import itertools
nouns = ['bacon', 'cheese', 'eggs', 'milk', 'fish', 'houses', 'dog']
CC = ['and', 'or']
def replacer_factory():
def create_permutations(match):
group1_string = (match.group(1)[:-1]) # strips trailing whitespace
# creates list of matched.group() with word 'and' or 'or' removed
nouns2 = filter(None, re.split(r',\s*', group1_string)) + [match.group(3)]
perm_nouns2 = list(itertools.permutations(nouns2))
CC_match = match.group(2) # this either matches word 'and' or 'or'
# create list that holds the permutations created in for loop below
perm_list = []
for comb in itertools.permutations(nouns2):
comb_len = len(comb)
if comb_len == 2:
perm_list.append(' '.join((comb[0], CC_match, comb[-1])))
elif comb_len == 3:
perm_list.append(', '.join((comb[0], comb[1], CC_match, comb[-1])))
elif comb_len == 4:
perm_list.append(', '.join((comb[0], comb[1], comb[2], CC_match, comb[-1])))
# does the match.group contain word 'and' or 'or'
if (match.group(2)) == "and":
joined = '*'.join(perm_list)
strip_comma = joined.replace("and,", "and")
completed = '|'+strip_comma+'|'
return completed
elif (match.group(2)) == "or":
joined = '*'.join(perm_list)
strip_comma = joined.replace("or,", "or")
completed = '|'+strip_comma+'|'
return completed
return create_permutations
def search_and_replace(text):
# use'nouns' and 'CC' lists to find a noun list phrase
# e.g 'bacon, eggs, and milk' is 1 example of a match
noun_patt = r'\b(?:' + '|'.join(nouns) + r')\b'
CC_patt = r'\b(' + '|'.join(CC) + r')\b'
patt = r'((?:{0},? )+){1} ({0})'.format(noun_patt, CC_patt)
replacer = replacer_factory()
return re.sub(patt, replacer, text)
def main():
with open('test_sentence.txt') as input_f:
read_f = input_f.read()
with open('output.txt', 'w') as output_f:
output_f.write(search_and_replace(read_f))
if __name__ == '__main__':
main()
“ test_sentence.txt”的內容:
I am 2 list with 'or': eggs or cheese.
I am 2 list with 'and': milk and eggs.
I am 3 list with 'or': cheese, bacon, and eggs.
I am 3 list with 'and': bacon, milk and cheese.
I am 4 list: milk, bacon, eggs, and cheese.
I am 5 list, I don't match.
I am 3 list with non match noun: cheese, bacon and pie.
因此,代碼都很好用,但是我遇到了一個局限性,我不知道該怎么解決。 此限制包含在for循環中。 就目前而言,我只創建了'if'和'elif'語句,它們僅達到elif comb == 4:
我實際上希望elif comb == 5:
變得不受限制,然后轉到elif comb == 5:
, elif comb == 6:
, elif comb == 7:
(嗯,在實際中,我真的不需要超越elif comb == 20
,但是要點是一樣的,我想允許這種可能性)。 但是,創建這么多的“ elif”語句並不現實。
關於如何解決這個問題的任何想法?
請注意,此處的“ test_sentence.txt”和變量“ noun”的列表僅是示例。 我實際上的“名詞”列表在1000左右,我將使用“ test_sentence.txt”中包含的較大文檔。
干杯達倫
附言-我努力為這個找到合適的標題!
如果您注意到了, if-elif
語句中的每一行都遵循大致相同的結構:首先獲取comb
列表中除最后一個元素之外的所有元素,添加CC_match
,然后添加最后一項。
如果我們將其寫為代碼,則會得到如下內容:
head = list(comb[0:-1])
head.append(CC_match)
head.append(comb[-1])
perm_list.append(', '.join(head))
然后,在您的for
循環內,您可以替換if-elif
語句:
for comb in itertools.permutations(nouns2):
head = list(comb[0:-1])
head.append(CC_match)
head.append(comb[-1])
perm_list.append(', '.join(head))
您還應該考慮添加一些錯誤檢查,以使如果comb
列表的長度等於零或一,程序不會產生奇怪的反應。
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