Why my regex replaces match pattern with function object rather than the replacement value?

Question

I need a regex that takes the replacement pattern from a dictionary. I wrote the following code but the pattern gets replaced by the function object rather than the replacement value.

>>> import re
>>> d1 = {'a':'1'}
>>> d2 = {'w':'2','k':'2'}
>>> re.sub(r'(a)([wk])', '%s%s' %(lambda m:d1[m.group()[0]], lambda m:d2[m.group()[1]]), 'waka')
'w<function <lambda> at 0x7f1e281efc08><function <lambda> at 0x7f1e281efcf8>a'

Expected output is 'w12a'.

falsetru · Accepted Answer · 2015-06-27 02:06:43Z

1

Becasue:

>>> import re
>>> '%s%s' %(lambda m:d1[m.group()[0]], lambda m:d2[m.group()[1]])
'<function <lambda> at 0x0000000002AB0128><function <lambda> at 0x0000000002AB0198>'

You're passing the above string as a replacement string.

Use replacement function as follow:

>>> import re
>>> d1 = {'a':'1'}
>>> d2 = {'w':'2','k':'2'}
>>> re.sub(r'(a)([wk])',
...        lambda m: d1[m.group(1)] + d2[m.group(2)],
...        'waka')
'w12a'

answered Jun 27, 2015 at 2:06

falsetru

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2 Comments

Irshad Bhat Over a year ago

Is there any way the above regex can be obtained without lambda.

falsetru Over a year ago

@BHATIRSHAD, With my knowledge, it's not possible unless it simple translation of a to 1, w to 2, k to 2. For such simple case, you don't need to use regular expression. Use string.translate

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