Azzera filtri
Azzera filtri

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 Commenti
Martin
Martin il 30 Mag 2021
Thanks for answering. join() and isfinite() is however a bit messy with regards to the cell piece.
dpb
dpb il 30 Mag 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 il 30 Mag 2021
Modificato: Stephen23 il 31 Mag 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 Commenti
Rik
Rik il 31 Mag 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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