I have a bunch of csv 's in folder.I want to combine them in an order.

I have the csv files as


Each of the csv has one line without no header, so the content of a state_i_result.csv will be like:


I want to combine in such a order that first line will be state_1_result_6.csv and second line will be state_1_result_8.csv and so on.

I tried with

cat state*.csv > final.csv

But it is not merging in a order.

  • is there any pattern on the is from state_i_result.csv? Every 2 starting from 6? So we can build a loop to generate them. – fedorqui Jan 4 '17 at 8:55
  • I edited the question now.The i are same in state_i_result_j.csv.The j has a pattern 6,8,10,12 only. – John Jan 4 '17 at 9:24

Since you know them, just loop through the list of indexes:

for i in 6 8 10 12;
   cat state_1_result_${i}.csv >> final.csv

You can also use a C-loop:

for ((i=6; i<=12; i+=2))
  • I am new to linux so how can I execute the first loop.I know I cannot directly type that in the command line.Could u please tell how to execute the for loop? – John Jan 4 '17 at 10:51
  • @John yes, you can type this in the command line. Just type the lines and when you find a new line, just type Enter! – fedorqui Jan 4 '17 at 10:53
  • Thank u I could not upvote because no enough reputations.Anyway thanks again – John Jan 4 '17 at 11:37
  • @John glad that it worked to you :) – fedorqui Jan 4 '17 at 11:40

It is good to see that the indices are just before the file extension.

cat state*.csv > final.csv

considers the sorting by ascii characters. That is, { 6, 8, 10, 12 } on sorting becomes { 10, 12 , 6, 8 }.

In the manual page of ls, there is an option -v, that sorts according to the version number, that is what is required. Therefore, you may use

cat `ls -v | grep .csv` > final.csv

if those are the only csv files in the directory.


or more simply

ls -v *.csv | xargs cat >> final.csv

-v order by version number

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