json char with too many decimals, need removal

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Martin
Martin am 29 Mai 2021
Kommentiert: Rik am 31 Mai 2021
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 Kommentare
Martin
Martin am 30 Mai 2021
Thanks for answering. join() and isfinite() is however a bit messy with regards to the cell piece.
dpb
dpb am 30 Mai 2021
I knew Stephen or similar would be along...
You'll note I specifically did NOT say it was elegant... :)

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Stephen23
Stephen23 am 30 Mai 2021
Bearbeitet: Stephen23 am 31 Mai 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 Kommentare
Martin
Martin am 30 Mai 2021
wow, thanks!
Rik
Rik am 31 Mai 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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