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I have a large number of CSV files, each file having multiple numeric columns. The first column is Unix timestamp. How can I sort such file? Thank you.
An example row of the file as below:
1376317246; 4; 11703.99824; 10477.029091173334; 89.51666666666667; 6.7108864E7; 2.0937962933333334E7; 0.0; 1.4; 235.53333333333333; 3.8666666666666667
I wouldn't call it a CSV file if it looks like that.
sort -t";" filename
should work. This will sort on all columns. If that's unappealing, then this should suit you.
From the docs:
-t, --field-separator=SEP use SEP instead of non-blank to blank transition
sort -t ';' -k2,2 to sort based on the second field. But not here we're you're sorting based on the full content of the line.
– Stéphane Chazelas
Mar 22 '19 at 14:42
; is larger than digits. So if there's a row whose first field is 137631724 (this is the value in the example without the trailing 6), sorting on the whole content will place this below the 1376317246, which would be incorrect.
– y_wc
Mar 22 '19 at 15:07
-t will make no difference. You'd need sort -t ';' -k1,1 for it to make a difference. But here, all you need and want is sort -n.
– Stéphane Chazelas
Sep 30 '19 at 13:37
you could use sort
sort --field-separator=';' --key=1 yourCSVfile.csv
if you need to create a new sorted file just add an output file:
sort --field-separator=';' --key=1 yourCSVfile.csv > sortedCSVfile.csv
I would use sort with following option:
-g (--general-numeric-sort) for numeric sort just in case, because default sorting mechanism is alphanimeric and sorting {9..11} would result in 10 11 9.sort -g filename
