如何计算每个文件中某一列中某个字符串出现的次数，结果文件名和 output 和 count + awk

Question

How do I count the number of occurrences of a string in a column in each file, and output the result filename and count + awk

I have these 2 files, :

>cat file.csv
col1,col2,col3,col4
col1,col2,col3
col1,col2
col1
col1,col2,col3,col4,col5


> cat fild.csv
col1,col2,col3,col4
col1,col2,col3
col1,col2
col1
col1,col2,col3,col4,col5

how do I get this output:(basically count the number of occurrences of a string (eg "col1")in a column in each file )

file.csv,5
fild.csv,5

Below, my attempts, for my reference:

Output column/field1

> awk -F, '$1 =="col1" {print $1}' file.csv
col1
col1
col1
col1
col1

Output filenamecolumn/field1, how do I add a comma as separator

> awk -F, '$1 =="col1" {print FILENAME $1}' file.csv
file.csvcol1
file.csvcol1
file.csvcol1
file.csvcol1
file.csvcol1

output Id like

file.csv,5

attempt working on 2 files:

> awk -F, '$1 =="col1" {print FILENAME $1}' fil*.csv
fild.csvcol1
fild.csvcol1
fild.csvcol1
fild.csvcol1
fild.csvcol1
file.csvcol1
file.csvcol1
file.csvcol1
file.csvcol1
file.csvcol1

But the output i'd like is this:

file.csv,5
fild.csv,5

Answer

this works for me:

awk 'BEGIN{FS=OFS=","} $1 == "col1" {cnt++} ENDFILE{print FILENAME, (cnt>0&&cnt?cnt:"0"); cnt=0}' fil*.csv
fild.csv,5
file1.csv,0
file.csv,5

Answer 1

If you're using GNU awk another potential solution is to use the ENDFILE special pattern, eg using @markp-fuso's example data:

cat filb.csv            # empty
cat filc.csv
col1

cat fild.csv
col1,col2,col3,col4
col1,col2,col3
col1,col2
col1
col1,col2,col3,col4,col5

cat file.csv
col1,col2,col3,col4
col1,col2,col3
col1,col2
col1
col1,col2,col3,col4,col5


awk 'BEGIN{FS=OFS=","} $1 == "col1" {cnt++} ENDFILE{print FILENAME, (cnt>0&&cnt?cnt:"0"); cnt=0}' fil*.csv
filb.csv,0
filc.csv,1
fild.csv,5
file.csv,5

# the 'cnt>0&&cnt?cnt:"0"' is to handle empty files
# basically, if there are lines print cnt otherwise, if
# there are no lines print "0"

Edit

As commented by @EdMorten, cnt+0 can be used instead of cnt>0&&cnt?cnt:"0" to handle empty files (much easier to remember,). eg

awk 'BEGIN{FS=OFS=","} $1 == "col1" {cnt++} ENDFILE{print FILENAME, cnt+0; cnt=0}' fil*.csv
filb.csv,0
filc.csv,1
fild.csv,5
file.csv,5

Answer 2

Adding a couple more files to the mix:

$ cat filb.csv            # empty
$ cat filc.csv
col2

One awk approach:

awk -v str='col1' '                       # pass in string to search for
BEGIN { FS=OFS=","
        for (i=1;i<ARGC;i++)
            count[ARGV[i]]=0              # initialize counter for all files; addresses issue where a file may not have any matches or file is empty (ie, count==0)
      }
      { for (i=1;i<=NF;i++)               # loop through fields looking for a match and ...
            if ($i==str)                  # if found then ...
               count[FILENAME]++          # increment our counter
      }
END   { for (fname in count)
            print fname,count[fname]
      }
' fil?.csv

This generates:

file.csv,5
filb.csv,0
fild.csv,5
filc.csv,0

NOTES:

• $i==str - assumes we're looking for an exact match on the value in the field (as opposed to a substring of the field's value)
• assumes we need to match str on any field/column in the file, otherwise we'll need to add an additional input variable to designate which column(s) to search
• output ordering is not guaranteed; OP can pipe the results to sort , or add some code to allow awk to sort the output before printing to stdout

---

An alternative grep|tr idea:

$ grep -oc "$str" fil?.csv | tr ':' ','
filb.csv,0
filc.csv,0
fild.csv,5
file.csv,5