SSH - 1 month ago 6x

Python Question

How would I limit match/replacement the leading zeros in e004_n07? However, if either term contains all zeros, then I need to retain one zero in the term (see example below). For the input string, there will always be 3 digits in the first value and 2 digits in the second value.

Example input and output

`e004_n07 #e4_n7`

e020_n50 #e20_n50

e000_n00 #e0_n0

Can this be accomplished with re.sub alone, or do I need to use re.search/re.match?

Answer

If you want to only remove zeros after letters, you may use:

```
([a-zA-Z])0+
```

Replace with `\1`

backreference. See the regex demo.

The `([a-zA-Z])`

will capture a letter and `0+`

will match 1 or more zeros.

```
import re
s = 'e004_n07'
res = re.sub(r'([a-zA-Z])0+', r'\1', s)
print(res)
```

Note that ** re.sub** will find and replace all non-overlapping matches (will perform a global search and replace). If there is no match, the string will be returned as is, without modifications. So, there is no need using additional

`re.match`

/`re.search`

.**UDPATE**

To keep 1 zero if the numbers only contain zeros, you may use

```
import re
s = ['e004_n07','e000_n00']
res = [re.sub(r'(?<=[a-zA-Z])0+(\d*)', lambda m: m.group(1) if m.group(1) else '0', x) for x in s]
print(res)
```

See the Python demo

Here, `r'(?<=[a-zA-Z])0+(\d*)'`

regex matches one or more zeros (`0+`

) that are after an ASCII letter (`(?<=[a-zA-Z])`

) and then any other digits (0 or more) are captured into Group 1 with `(\d*)`

. Then, in the replacement, we check if Group 1 is empty, and if it is empty, we insert `0`

(there are only zeros), else, we insert Group 1 contents (the remaining digits after the first leading zeros).

Source (Stackoverflow)

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