return string with first match Regex

Question

I want to get the first match of a regex   In this case  I got a list   text    aa33bbb44  re findall   d   text         33    44     I could extract the first element of the list   text    aa33bbb44  re findall   d   text  0        33    But that only works if there is at least one match  otherwise I ll get an error   text    aazzzbbb  re findall   d   text  0       IndexError  list index out of range   In which case I could define a function   def return first match text       try          result   re findall   d   text  0      except Exception  IndexError          result          return result   Is there a way of obtaining that result without defining a new function

User · Answer

You shouldn t be using  findall   at all -  search   is what you want   It finds the leftmost match  which is what you want  or returns None if no match exists    m   re search pattern  text  result   m group 0  if m else      Whether you want to put that in a function is up to you   It s unusual to want to return an empty string if no match is found  which is why nothing like that is built in   It s impossible to get confused about whether  search   on its own finds a match  it returns None if it didn t  or an SRE Match object if it did

User · Answer

You can do   x   re findall   d    text  result   x 0  if len x   gt  0 else      Note that your question isn t exactly related to regex  Rather  how do you safely find an element from an array  if it has none

User · Answer

You could embed the    default in your regex by adding       gt  gt  gt  re findall   d       aa33bbb44   0   33   gt  gt  gt  re findall   d       aazzzbbb   0      gt  gt  gt  re findall   d          0       Also works with re search pointed out by others    gt  gt  gt  re search   d       aa33bbb44   group    33   gt  gt  gt  re search   d       aazzzbbb   group       gt  gt  gt  re search   d          group

User · Answer

Maybe this would perform a bit better in case greater amount of input data does not contain your wanted piece because except has greater cost    def return first match text       result   re findall   d   text      result   result 0  if result else        return result

User · Answer

I d go with   r   re search   d    ch  result   return r group 0  if r else      re search only looks for the first match in the string anyway  so I think it makes your intent slightly more clear than using findall

User · Answer

If you only need the first match  then use re search instead of re findall    gt  gt  gt  m   re search   d     aa33bbb44    gt  gt  gt  m group    33   gt  gt  gt  m   re search   d     aazzzbbb    gt  gt  gt  m group   Traceback  most recent call last     File   lt pyshell 281 gt    line 1  in  lt module gt      m group   AttributeError   NoneType  object has no attribute  group    Then you can use m as a checking condition as    gt  gt  gt  m   re search   d     aa33bbb44    gt  gt  gt  if m          print  First number found       format m group         else          print  Not Found     First number found   33

[python] return string with first match Regex

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