Python regular expression substitution function using lambda in the replacement field

Question

I am trying to mimic 'i am your'.title() string function with the following code to get the output as I Am Your. But it gives the same i am your as output without capitalizing the words

re.sub(r"\b'\w", lambda m: m.group().upper(), 'i am your')

During the process of verfiying it, I came across this following snippet in the internet but not able to comprehend how the lambda gets the input from/as:

re.sub("[ab]", lambda x: x.group(0).upper(), "charly") # chArly

because,

>>> f = lambda x: x.group(0).upper()
>>> f("[ab]")
AttributeError: 'str' object has no attribute 'group'

>>> f(["a","b"])
AttributeError: 'list' object has no attribute 'group'

To understand it better, I tried to decode it with Python disassembler, but still it seems vague

>>> dis.dis('re.sub("[ab]", lambda x: x.group(0).upper(), "charly")')
          0 POP_JUMP_IF_FALSE 11877
          3 POP_JUMP_IF_TRUE 25205
          6 STORE_SLICE+0
          7 <34>
          8 DELETE_NAME     25185 (25185)
         11 FOR_ITER        11298 (to 11312)
         14 SLICE+2
         15 IMPORT_NAME     28001 (28001)
         18 DELETE_GLOBAL   24932 (24932)
         21 SLICE+2
         22 SETUP_LOOP       8250 (to 8275)
         25 SETUP_LOOP      26414 (to 26442)
         28 POP_JUMP_IF_FALSE 30063
         31 JUMP_IF_TRUE_OR_POP 12328
         34 STORE_SLICE+1
         35 <46>
         36 <117>           28784
         39 LOAD_NAME       10354 (10354)
         42 STORE_SLICE+1
         43 <44>
         44 SLICE+2
         45 <34>
         46 DUP_TOPX        24936
         49 POP_JUMP_IF_FALSE 31084
         52 <34>
         53 STORE_SLICE+1

I would use regular str functions instead of regex: ' '.join(s[:1].upper() + s[1:] for s in 'i am your'.split()). — Olvin Roght
– Olvin Roght, Commented Oct 2, 2020 at 14:26

Liju · Accepted Answer · 2020-10-02 15:03:24Z

1

You need to use upper() to get the first char in caps
There is a ' between word break and word escape literals, you need to remove that.
And you need to use m.group(0) or m[0] to get matched string.

>>> re.sub(r"\b\w",lambda m: m[0].upper(),"i am your")
'I Am Your'

answered Oct 2, 2020 at 15:03

Liju

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4 Comments

Ibrahim Quraish Over a year ago

Ah, silly mistakes! :) Thank you Liju. I replaced lower() to upper() in my original posting and left others as it is. Between, I noticed m.group(0) works in py2.7 and m[0] works in py3

Ibrahim Quraish Over a year ago

Liju or anyone, can please explain how the lambda works in the 2nd code snippet of my post? I have made the question bold.

Liju Over a year ago

Because you have given 0 in group() which you didn't do in first snippet.

Ibrahim Quraish Over a year ago

Suppose if I write lambda function this way, f = lambda x: x.group(0).upper() then how to pass the argument pattern and the source string to match the pattern with n this case?

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