我正在嘗試在一串 PL/FOL 公式中構(gòu)建文字列表，并且相關(guān)的代碼正在查找匹配項(xiàng)，但將它們作為空白返回。我試過re.escape(formula)了，什么也沒做。我也嘗試過該findall模式的簡單變體，但它們隨后會(huì)生成空列表。def clean(formula): formula = formula.strip() formula = re.sub("\( +", "(", formula) formula = re.sub(" +\)", ")", formula) formula = re.sub("(?P<b_ops>[&v→?])", " " + "\g<b_ops>" + " ", formula) formula = re.sub("[ ]+", " ", formula) # Make an inventory of literals for the original formula. orig_lit_inv = re.findall("[~]*[A-Z]([a-u]|[w-z]){0,}", formula) print(orig_lit_inv)this_WFF = "(P) & ~(~(Q → (R & ~S)))"clean(formula=this_WFF)當(dāng)我打印結(jié)果時(shí)，我得到['', '', '', '']. 換句話說，它正在查找匹配項(xiàng)，但返回空白字符串作為匹配項(xiàng)，此時(shí)它至少應(yīng)該返回[A-Z]. 以this_WFF作為參數(shù)，clean(formula)應(yīng)該打印['P', 'Q', 'R', '~S'].
1 回答
鳳凰求蠱
TA貢獻(xiàn)1825條經(jīng)驗(yàn) 獲得超4個(gè)贊
如果模式中存在一個(gè)或多個(gè)捕獲組，則返回組列表；如果模式有多個(gè)組，這將是一個(gè)元組列表。
您的正則表達(dá)式包含一個(gè)捕獲組，因此findall永遠(yuǎn)不會(huì)為[A-Z]正則表達(dá)式返回任何內(nèi)容。更改([a-u]|[w-z])為(?:[a-u]|[w-z])查看差異：
>>> this_WFF = "(P) & ~(~(Q → (R & ~S)))"
>>> def clean(formula):
... formula = formula.strip()
... formula = re.sub("\( +", "(", formula)
... formula = re.sub(" +\)", ")", formula)
... formula = re.sub("(?P<b_ops>[&v→?])", " " + "\g<b_ops>" + " ", formula)
... formula = re.sub("[ ]+", " ", formula)
... # Make an inventory of literals for the original formula.
... orig_lit_inv = re.findall("[~]*[A-Z]([a-u]|[w-z]){0,}", formula)
... print(orig_lit_inv)
...
>>> clean(this_WFF)
['', '', '', '']
>>> def clean(formula):
... formula = formula.strip()
... formula = re.sub("\( +", "(", formula)
... formula = re.sub(" +\)", ")", formula)
... formula = re.sub("(?P<b_ops>[&v→?])", " " + "\g<b_ops>" + " ", formula)
... formula = re.sub("[ ]+", " ", formula)
... # Make an inventory of literals for the original formula
... orig_lit_inv = re.findall("[~]*[A-Z](?:[a-u]|[w-z]){0,}", formula)
... print(orig_lit_inv)
...
>>> clean(this_WFF)
['P', 'Q', 'R', '~S']
由于現(xiàn)在正則表達(dá)式不包含捕獲組findall，因此只需在結(jié)果中返回“組 0”（即整個(gè)匹配項(xiàng)）的內(nèi)容。
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