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How to print values that are greater than 0.1 in at least 80% of the samples?


 
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Top Forums Shell Programming and Scripting How to print values that are greater than 0.1 in at least 80% of the samples?
# 1  
Old 02-17-2015
How to print values that are greater than 0.1 in at least 80% of the samples?

input
Code:
        sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
v1    0.2     0.1     0.1    0       1       2       3       4       9       10
v2    0       0       0.01    0       0       0       0       0       0       0
v3    0       0       0       0       0       0       0       0       0       0
v4    0.2     0       0       0       0       0       0       0       0       0
v5    0.1     0.1     0.2     0.2     10      2       3       5       6       7

output
Code:
sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
v5    0.1     0.1     0.2     0.2     10      2       3       5       6       7

Tried
I tried sth that able to print values that are greater than 0.1
Code:
 awk '{for (i=2;i<=NF;i++) if ($i>0.1) {print $0;next} }'

# 2  
Old 02-17-2015
Code:
awk '{ T=0 ; for (i=2;i<=NF;i++) if ($i>0.1) {T++ }  if(T > (NF * 0.8)) print}'

# 3  
Old 02-17-2015
@corona688: it is just printing the header. I think something is wrong ?
# 4  
Old 02-17-2015
Try
Code:
awk '{L=1; for (i=2;i<=NF;i++) L=L*($i>=0.1)} L' file4
        sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
v5    0.1     0.1     0.2     0.2     10      2       3       5       6       7

Please be aware that the header is printed "by sheer coincidence". To make it print safely, add sth like NR==1; or so...

@Corona688: I'm not sure I get your logics. Could you explain?

Last edited by RudiC; 02-17-2015 at 12:10 PM.. Reason: Removed my silly question to Corona688
# 5  
Old 02-17-2015
Hello quincyjones,

Could you please try following and let me know if this helps.(Little addition to Corona's code)
Code:
awk 'BEGIN{ T=0} ; {if(NR==1){print $0} else if(NR>1){for (i=2;i<=NF;i++) if ($i>0.1) {T++ }  if(T > (NF * 0.8)) {print;T=""}}}'   Input_file

Output will be as follows.
Code:
        sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
v5    0.1     0.1     0.2     0.2     10      2       3       5       6       7

Thanks,
R. Singh
This User Gave Thanks to RavinderSingh13 For This Post:
# 6  
Old 02-17-2015
Rats! I didn't read the title ... be back soon ...

---------- Post updated at 17:06 ---------- Previous update was at 16:57 ----------

... adapting Corona688's proposal slightly (as there are 10 date fields but 11 in total; and the request was "at least 80%"):
Code:
awk     '{T=0; for (i=2;i<=NF;i++) T+=($i>0.1)}
         T >= ((NF-1) * 0.8)
        ' file
        sample1 sample2 sample3 sample4 sample5 sample6 sample7 sample8 sample9 sample10
v5    0.1     0.1     0.2     0.2     10      2       3       5       6       7


Last edited by RudiC; 02-17-2015 at 12:13 PM..
# 7  
Old 02-17-2015
Thank you all. it's working great. However could someone please explain the logic behind
Quote:
if (T>NF*0.8)
.

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