Effacer les filtres
Effacer les filtres

json char with too many decimals, need removal

3 vues (au cours des 30 derniers jours)
Martin
Martin le 29 Mai 2021
Commenté : Rik le 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 commentaires
Martin
Martin le 30 Mai 2021
Thanks for answering. join() and isfinite() is however a bit messy with regards to the cell piece.
dpb
dpb le 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 le 30 Mai 2021
Modifié(e) : Stephen23 le 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 commentaires
Martin
Martin le 30 Mai 2021
wow, thanks!
Rik
Rik le 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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