Remove markdown code block from python string

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I'm trying to remove markdown code blocks from the following string:

Problem with encoding a very large BigInteger or BigDecimal without fraction

We have an issue to encode a very large BigDecimal. For example, when I tried to encode _7.533938258014959827307132527342E+545_ I got the following error:

``` java
Error com.n1analytics.paillier.EncodeException: Input value cannot be encoded.
at com.n1analytics.paillier.StandardEncodingScheme.encode(StandardEncodingScheme.java:115)
at com.n1analytics.paillier.StandardEncodingScheme.encode(StandardEncodingScheme.java:239)
...
```

The reason why I got the error is as follows:

I've tried:

txt = re.sub('(```[a-z]*\n[\s\S]*?\n```)', '', txt) # https://regex101.com/r/aA5bI3/3
txt = re.sub('(```.+?```)', '', txt)  # https://regex101.com/r/aA5bI3/3

Even though I saw some of these regex at https://coderwall.com/p/r6b4xg/regex-to-match-github-s-markdown-code-blocks and https://regex101.com/r/aA5bI3/3, I had no success

python regex
2 Answers
user_7870160

The first pattern almost works, but in the example data there is a space before java

If the spaces and the chars a-z are the only acceptable characters after the backticks and they are optional, you could match optional whitespace chars without a newline [^\S\r\n]*

The pattern could look like

```[^\S\r\n]*[a-z]*\n.*?\n```

Instead of using [\s\S]*? you can use the re.DOTALL flag to make the dot match a newline.

Regex demo | Python demo

For example

txt = re.sub(r"```[^\S\r\n]*[a-z]*\n.*?\n```", '', txt, 0, re.DOTALL)

If the backticks always begin at the start of the string, and also end at the start of the string, you could make the pattern a bit more efficient matching all the lines in between that do not start with 3 backticks using a negative lookahead to prevent unnecessary backtracking.

As the pattern uses an anchor, you should use the re.MULTILINE flag.

^```[^\S\r\n]*[a-z]*(?:\n(?!```$).*)*\n```

Explanation

  • ^ Start of string
  • ``` Match 3 backticks
  • [^\S\r\n]*[a-z]* Match optional spaces without a newline and optional chars a-z
  • (?: Non capture group
    • \n(?!```$) Match a newline and assert that the line does not start with 3 backticks
    • .* If that is the case, match the whole line
  • )* Close non capture group and repeat 0+ times to match all lines
  • \n``` Match a newline and 3 backticks

Regex demo | Python demo

For example

txt = re.sub(r"^```[^\S\r\n]*[a-z]*(?:\n(?!```$).*)*\n```", '', txt, 0, re.MULTILINE)
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