Rounding numbers of big data

#1
Hi,
I'm new here. There is a difficulty (I'm not professional) about a large amount of data.
For example,

0.00137716715090196 -0.000106048713201803 -0.00231457180318346 -0.00486321687763309 -0.00583302412031425 -0.00438296123102898 -0.00156538780467394 0.000205951890463430 0.000693199703720858
0.00429909913296669 0.00124640829884964 -0.00297989682574049 -0.00639892235240998 -0.00611161038482716 -0.00320670857869514 0.000115201223084927 0.00190628176535631 0.00166190728156096
0.00651323662004019 0.00142996126754939 -0.00277733860291394 -0.00473932564078852 -0.00404290798224363 -0.00186799033729221 2.18837962138496e-05 0.00102999265990287 0.00248534049241512
0.00748493375431313 0.00207868010154096 -0.000215118689763324 -0.000789173559639632 -0.000639549549550765 -0.000926601851903136 -0.00133986243163238 -8.16996449505892e-05 0.00361159476944932
0.00942490138175875 0.00513733038309198 0.00427744669519739 0.00397550134456293 0.00277264970424855 0.000398919221571838 -0.00152267825421809 0.000657255438959982 0.00544276290718545
0.0139719586505275 0.0110420210687043 0.00930847882384559 0.00755637424453657 0.00483155749897842 0.00248986428153341 0.00192258531702006 0.00379083045439589 0.00745059436094635

but it's just a small part of data; my question is that: how do I round to four decimals without blurring it? Like 0.00012 is different from 0.00011 but It will be round to 0.0001; I don't want it. I heard a method that rounding them within a 95% confidence interval but I don't know the details. Is that I have to add two decimals to make them within 95%CI? Hope you could explain to me. Thank you so much.
 

vinux

Dark Knight
#2
Changing the unit (multiplying by 1000 or 10000) and rounding the value may be appropriate here. If the maximum value is a small number (<0.1), you can multiply with the appropriate 10^x and round the number.
 

Dason

Ambassador to the humans
#3
I don't really understand the request. You want to round but make sure the resulting values are unique or something? Please explain your motivation.
 

AngleWyrm

Active Member
#5
The posted sample data
Untitled.png
The chart on the left shows a y axis displayed with a resolution of 1/10000
The chart on the right plots the full value x vs int(10000*x)/10000 and the resulting trendline has a slope of 1