space delimited file handling - regex

Question

I have insider transactions of a company in a space delimited file. Sample data looks like the following:
1 Gilliland Michael S January 2,2013 20,000 19
2 Still George J Jr January 2,2013 20,000 19
3 Bishkin S. James February 1,2013 150,000 21
4 Mellin Mark P May 28,2013 238,000 25.26
Col1 is Serial# that I dont need to print
Col2 is the name of the person who did trades. This column is not consistent. It has first name and second name and middle initial and for some insiders salutations as well (Mr, Dr. Jr etc)
col3 is the date format Month Day,Year
col4 is the number of shares traded
col5 is the price at which shares were purchased or sold.
I need you guys help to print each column value separately. Thanks for your help.

Answer 1

Count the total number of fields read; the difference between that and the number of non-name fields gives you the width of the name.
#!/bin/bash
# uses bash features, so needs a /bin/bash shebang, not /bin/sh
# read all fields into an array
while read -r -a fields; do
# calculate name width assuming 5 non-name fields
name_width=$(( ${#fields[#]} - 5 ))
cur_field=0
# read initial serial number
ser_id=${fields[cur_field]}; (( ++cur_field ))
# read name
name=''
for ((i=0; i<name_width; i++)); do
name+=" ${fields[cur_field]}"; (( ++cur_field ))
done
name=${name# } # trim leading space
# date spans two fields due to containing a space
date=${fields[cur_field]}; (( ++cur_field ))
date+=" ${fields[cur_field]}"; (( ++cur_field ))
# final fields are one span each
num_shares=${fields[cur_field]}; (( ++cur_field ))
price=${fields[cur_field]}; (( ++cur_field ))
# print in newline-delimited form
printf '%s\n' "$ser_id" "$name" "$date" "$num_shares" "$price" ""
done
Run as follows (if you saved the script as process):
./process <input.txt >output.txt

Answer 2

It might be a little easier in perl.
perl -lane '
#date = splice #F, -4, 2;
#left = splice #F, -2, 2;
splice #F, 0, 1;
print join "|", "#F", "#date", #left
' file
Gilliland Michael S|January 2,2013|20,000|19
Still George J Jr|January 2,2013|20,000|19
Bishkin S. James|February 1,2013|150,000|21
Mellin Mark P|May 28,2013|238,000|25.26
You can change the delimiter in the join as per your requirement.

Answer 3

Here is the data separated using awk
awk '{c1=$1;c5=$NF;c4=$(NF-1);c3=$(NF-3)FS$(NF-2);$1=$NF=$(NF-1)=$(NF-2)=$(NF-3)="";gsub(/^ | *$/,"");c2=$0;print c1"|"c2"|"c3"|"c4"|"c5}' file
1|Gilliland Michael S|January 2,2013|20,000|19
2|Still George J Jr|January 2,2013|20,000|19
3|Bishkin S. James|February 1,2013|150,000|21
4|Mellin Mark P|May 28,2013|238,000|25.26
You know have your data in variable c1 to c5
Or better displayed here:
awk '{c1=$1;c5=$NF;c4=$(NF-1);c3=$(NF-3)FS$(NF-2);$1=$NF=$(NF-1)=$(NF-2)=$(NF-3)="";gsub(/^ | *$/,"");c2=$0;print c1"|"c2"|"c3"|"c4"|"c5}' file | column -t -s "|"
1 Gilliland Michael S January 2,2013 20,000 19
2 Still George J Jr January 2,2013 20,000 19
3 Bishkin S. James February 1,2013 150,000 21
4 Mellin Mark P May 28,2013 238,000 25.26