Adam - 1 year ago 41

Python Question

How to capture

`[0-9]+`

`(\], \[)`

`[[[[u'1', u'2'], u'3'], u'4'], [[[u'1', u'2'], u'4'], [[u'1', u'5'], u'4']]]`

I would like to capture three groups,

`1 2 3 4`

`1 2 4`

`1 5 4`

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Answer Source

Assuming that you don't have sub-patterns like `[[u'1', u'2'], [u'3',u'5']]`

(multiple nested sub-groups at the same level, in which case you need to use a stack and parse like pushdown automata) you could do this with regular expressions in two steps:

(1) split the expression with regex `\]\s*,\s*\[`

to get the groups first, you will get 3 groups for the example provided.

(2) within each group use the regex `[^0-9u]*u'([0-9]+)'[^0-9u]*`

to extract the digits.

For example, in `R`

, the code will be:

```
str <- "[[[[u'1', u'2'], u'3'], u'4'], [[[u'1', u'2'], u'4'], [[u'1', u'5'], u'4']]]"
groups <- unlist(strsplit(str, split='\\]\\s*,\\s*\\['))
pattern <- "[^0-9u]*u'([0-9]+)'[^0-9u]*"
lapply(groups, function(str) gsub(pattern, "\\1", regmatches(str,gregexpr(pattern,str))[[1]]))
#[[1]]
#[1] "1" "2" "3" "4"
#[[2]]
#[1] "1" "2" "4"
#[[3]]
#[1] "1" "5" "4"
```

In `python`

:

```
import re
str = "[[[[u'1', u'2'], u'3'], u'4'], [[[u'1', u'2'], u'4'], [[u'1', u'5'], u'4']]]"
groups = re.split('\]\s*,\s*\[', str)
pattern = "[^0-9u]*u'([0-9]+)'[^0-9u]*"
print map(lambda x: re.findall(pattern, x), groups)
# [['1', '2', '3', '4'], ['1', '2', '4'], ['1', '5', '4']]
```

you could map the digits to integers if required.

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