Bash：读取 CSV 文本文件并查找行的平均值

Question

This is the sample input (the data has user-IDs and the number of hours spent by the user):

Computer ID,User ID,M,T,W,T,F
Computer1,User3,5,7,3,5,2
Computer2,User5,8,8,8,8,8
Computer3,User4,0,8,0,8,4
Computer4,User1,5,4,5,5,8
Computer5,User2,9,8,10,0,0

I need to read the data, find all User-IDs ending in even numbers (2,4,6,8..) and find average number of hours spent (over five days).

I wrote the following script:

hoursarray=(0,0,0,0,0)
while IFS=, read -r col1 col2 col3 col4 col5 col6 col7 || [[ -n $col1 ]]
do
    if [[ $col2 == *"2" ]]; then
        #echo "$col2"
        ((hoursarray[0] = col3 + col4 + col5 + col6 + col7))
    elif  [[ $col2 == *"4" ]]; then 
        #echo "$col2"
        ((hoursarray[1] = hoursarray[1] + col3 + col4 + col5 + col6 + col7))
    elif [[ $col2 == *"6" ]]; then
        #echo "$col2"
        ((hoursarray[2] = hoursarray[2] + col3 + col4 + col5 + col6 + col7))
    elif [[ $col2 == *"8" ]]; then
        #echo "$col2"
        ((hoursarray[3] = hoursarray[3] + col3 + col4 + col5 + col6 + col7))
    elif [[ $col2 == *"10" ]]; then
        #echo "$col2"
        ((hoursarray[4] = hoursarray[4] + col3 + col4 + col5 + col6 + col7))
    fi
done < <(tail -n+2 user-list.txt)
echo ${hoursarray[0]}
echo "$((hoursarray[0]/5))"

This is not a very good way of doing this. Also, the numbers arent adding up correctly.

I am getting the following output (for the first one - user2):

27
5

I am expecting the following output:

27
5.4

What would be a better way to do it? Any help would be appreciated.

TIA

Answer 1

You issue is echo "$((hoursarray[0]/5))" Bash does not have floating point, so it returns the integer portion only.

Easy to demonstrate:

$ hours=27
$ echo "$((hours/5))"
5

If you want to stick to Bash, you could use bc for the floating point result:

$ echo "$hours / 5.0" | bc -l
5.40000000000000000000

Or use awk , perl , python , ruby etc.

Here is an awk you can parse out. Easily modified to you use (which is a little unclear to me)

awk -F, 'FNR==1{print $2; next} 
     {arr[$2]+=($3+$4+$5+$6+$7) }   
     END{ for (e in arr) print e "\t\t" arr[e] "\t" arr[e]/5 }' file 

Prints:

User ID
User1       27  5.4
User2       27  5.4
User3       22  4.4
User4       20  4
User5       40  8

If you only want even users, filter for User that end in any of 0,2,4,6,8:

awk -F, 'FNR==1{print $2; next} 
         $2~/[24680]$/ {arr[$2]+=($3+$4+$5+$6+$7) } 
         END{ for (e in arr) print e "\t\t" arr[e] "\t" arr[e]/5 }' file

Prints:

User ID
User2       27  5.4
User4       20  4

Answer 2

Your description is fairly imprecise, but here's an attempt primarily based on the sample output:

awk -F, '$2~/[24680]$/{for(i=3;i<=7;i++){a+=$i};print a;printf "%.2g\n",a/5; a=0}' file 
20
4
27
5.4

$2~/[24680]$/ makes sure we only look at "even" user-IDs.

for(i=3;i<=7;i++){} iterates over the day columns and adds them.

Edit 1: Accommodating new requirement:

awk -F, '$2~/[24680]$/{for(i=3;i<=7;i++){a+=$i};printf "%s\t%.2g\n",$2,a/5;a=0}' saad 
User4   4
User2   5.4

Answer 3

Sample data showing userIDs with even and odd endings, userID showing up more than once (eg, User2 ), and some non-integer values:

$ cat user-list.txt
Computer ID,User ID,M,T,W,T,F
Computer1,User3,5,7,3,5,2
Computer2,User5,8,8,8,8,8
Computer3,User4,0,8,0,8,4
Computer4,User1,5,4,5,5,8
Computer5,User2,9,8,10,0,0
Computer5,User120,9,8,10,0,0
Computer5,User2,4,7,12,3.5,1.5

One awk solution to find total hours plus averages, across 5x days, with duplicate userIDs rolled into a single set of numbers, but limited to userIDs that end in an even number:

$ awk -F',' 'FNR==1 { next } $2 ~ /[02468]$/ { tot[$2]+=($3+$4+$5+$6+$7) } END { for ( i in tot ) { print i, tot[i], tot[i]/5 } }' user-list.txt

Where:

• -F ',' - use comma as input field delimiter
• FNR==1 { next } - skip first line
• $2 ~ /[02468]$/ - if field 2 ends in an even number
• tot[$2]+=($3+$4+$5+$6+$7) - add current line's hours to array where userID is the array index; this will add up hours from multiple input lines (for same userID) into a single array cell
• for (...) { print ...} - loop through array indices printing the index, total hours and average hours (total divided by 5)

The above generates:

User120 27 5.4
User2 55 11
User4 20 4

Depending on OPs desired output the print can be replaced with printf and the desired format string ...

Answer 4

 Here is your script modified a little bit:
  
 while IFS=, read -r col1 col2 col3 || [[ -n $col1 ]]
 do
       (( $(sed 's/[^[:digit:]]*//' <<<$col2) % 2 )) || ( echo -n "For $col1 $col2 average is: " && echo "($(tr , + <<<$col3))/5" | bc -l )
 done < <(tail -n+2 list.txt)

prints:

 For Computer3 User4 average is: 4.00000000000000000000
 For Computer5 User2 average is: 5.40000000000000000000