I have the following sentence: "The size of the lunch box is around 1.5l or 1500ml"
How can I change this to: "The size of the lunch box is around 1.5 liter or 1500 milliliter"
In some cases, the value might also be present as "1.5 l or 1500 ml" with a space.
I am not be able to capture the "l" or "ml" when I am trying to build a function, or it is giving me an escape error.
I tried:
def stnd(text):
text = re.sub('^l%',' liter', text)
text = re.sub('^ml%',' milliliter', text)
text = re.sub('^\d+\.\d+\s*l$','^\d+\.\d+\s*liter$', text)
text = re.sub('^^\d+\.\d+\s*ml$%','^\d+\.\d+\s*milliliter$', text)
return text
You could use a dict to list all the units as the key, and use a pattern to find a digit followed by either ml
or l
which you could then use as the key for the dict to get the value.
(?<=\d)m?l\b
The pattern matches:
(?<=\d)
Positive lookbehind, assert a digit to the leftm?l\b
Match an optional m
followed by b and a word boundarySee a regex demo.
Example
s = "The size of the lunch box is around 1.5l or 1500ml"
pattern = r"(?<=\d)m?l\b"
dct = {
"ml": "milliliter",
"l": "liter"
}
result = re.sub(pattern, lambda x: " " + dct[x.group()] if x.group() in dct else x, s)
print(result)
Output
The size of the lunch box is around 1.5 liter or 1500 milliliter