json char with too many decimals, need removal

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Martin
Martin 2021-5-29
评论： Rik 2021-5-31
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
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Martin
Martin 2021-5-30
Thanks for answering. join() and isfinite() is however a bit messy with regards to the cell piece.
dpb
dpb 2021-5-30
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 2021-5-30
编辑：Stephen23 2021-5-31
Ran in:
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}'
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Rik
Rik 2021-5-31
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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