Python is learning a

The metacharacters. ^ $ * + ? {}. The match is: If a match is written to a point, if you match more than one word, write a few dots >>> a = re.findall (' b...dd ', ' BSSSDD ') >>> print (a) [' BSSSDD '] such as: > A = Re.findall (' B.. DD ', ' BSSSDD ') >>> print (a) [] explanation: Since the two points need to match the middle there are three causes that cannot match the ^ expression:>>> re.findall (' ^a. S ', ' abcsassss ') [' ABCs ']>>> re.findall (' A. S ', ' abcsassss ') [' ABCs ', ' asss ']>>> re.findall (' ^a. S ', ' abcsassss ') [' ABCs ']>>> if the comma contains a matching one, if ^ outside the comma is the maximum match content. The only thing that can be satisfied is to match it. The $ match is a match from the last start, if there is a hint empty, if there is a:>>> re.findall (' a.b$ ', ' Abbdddddabb ') [' ABB ']>>> re.findall (' $a. B ', ' Abbdddddabb ') []* match: Indicates: If the match succeeds, the last character will be the maximum to match the content:>>> re.findall (' abc* ', ' SABCCCC ') [' ABCCCC ']>>> Re.findall (' abc* ', ' SABCCCGEWEC ') [' ABCCC ']>>>? Expression: If the match succeeds stop matching;>>> re.findall (' abc? ', ' ABCCC ') [' abc ']>>> re.findall (' abc+ ', ' ABCCC ') [' ABCCC '] >>> re.findall (' abc* ', ' ABCCC ') [' ABCCC ']>>> re.findall (' abc? ', ' ABCCC ') [' abc ']{} match rule: Now conditions such as: {1,3} Indicates that only one of the above can be satisfied. The first number needs to be less than the second one otherwise it will go wrong >>> re.findall (' Abc{1,4} ', ' ABCCC ') [' ABCCC ']>>>>>> re.findall (' abc{1,4} ', ' ABCCCCCCCCCCCCCCCC ') [' ABCCCC ']>>> Re.findall (' abc{1,2} ', ' ABCCCCCCCCCCCCCCCC ') [' ABCC '] meta character appended character set []: Indicates that there are several values in the relationship, but only one, such as non-additive {to limit length matching}>>> Re.findall (' A[b,c]{1}d? ', ' abcddssaddddd ') [' AB ']>>>

Python is learning one