json char with too many decimals, need removal

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I got some json str = '{"hund": 0.3253533250000000000000000000000000, "kat": "dfsdfs", "baenkebider": 0.002021203321320000000000000000000000}';
In reality its a longer string with even more figures. I need to remove all those long decimals. 7-8 decimals is enough, so I end up with something like:
str = '{"hund": 0.3253533, "kat": "dfsdfs", "baenkebider": 0.0020212}';
If someone can help with an elegant solution I will appreciate it a lot!
Thanks in advance,
-best
mergh
  3 Comments
Martin
Martin on 30 May 2021
Thanks for answering. join() and isfinite() is however a bit messy with regards to the cell piece.
dpb
dpb on 30 May 2021
I knew Stephen or similar would be along...
You'll note I specifically did NOT say it was elegant... :)

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Accepted Answer

Stephen23
Stephen23 on 30 May 2021
Edited: Stephen23 on 31 May 2021
str = '{"hund": 0.3253533250000000000000000000000000, "kat": "dfsdfs", "baenkebider": 0.002021203321320000000000000000000000}';
Method one (truncate to 9 characters):
out = regexprep(str,'\d+\.\d+','${$&(1:9)}')
out = '{"hund": 0.3253533, "kat": "dfsdfs", "baenkebider": 0.0020212}'
Method two (seven fractional digits):
fun = @(s)sprintf('%.7f',sscanf(s,'%f'));
out = regexprep(str,'\d+\.\d+','${fun($&)}')
out = '{"hund": 0.3253533, "kat": "dfsdfs", "baenkebider": 0.0020212}'
  2 Comments
Rik
Rik on 31 May 2021
This will of course not work on an arbitrary JSON string, so you need to be careful if you want to do so. You could consider using a custom JSON encoder that allows you to trim trailing 0 in decimal notation.

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