您可以尝试使用awk

awk '
    BEGIN{FS=","; OFS=", "}
    !($1 in max) || $2>max[$1]{max[$1]=$2}
    !($1 in min) || $2<min[$1]{min[$1]=$2}
    END{
        for(k in max){print k, min[k], max[k]}
    }' input.txt

你会得到

Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116

另一种方法是先按column1排序，然后按column2排序，然后对每个任务取第一个和最后一个值，就像这样

awk -F, '{arr[$1]=arr[$1] $2} END {for(key in arr) print key, arr[key]}' <(sort -t 1 -k 1,2 file) | awk '{OFS=", "; print $1, $2, $NF}'

样品运行：

$ cat file 
Task2, 3421
Task3, 3300
Task1, 1000
Task2, 1100
Task3, 1200
Task3, 1209
Task4, 1299
Task3, 1289
Task1, 1389
Task2, 1211
Task5, 1216
Task2, 1416
Task1, 2100
Task6, 2416
Task5, 2216
Task7, 1116
$ sort -t 1 -k 1,2 file
Task1, 1000
Task1, 1389
Task1, 2100
Task2, 1100
Task2, 1211
Task2, 1416
Task2, 3421
Task3, 1200
Task3, 1209
Task3, 1289
Task3, 3300
Task4, 1299
Task5, 1216
Task5, 2216
Task6, 2416
Task7, 1116
$ awk -F, '{arr[$1]=arr[$1] $2} END {for(key in arr) print key, arr[key]}' <(sort -t 1 -k 1,2 file) | awk '{OFS=", "; print $1, $2, $NF}'
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116

使用gawk的array of arrays：

gawk 'BEGIN{OFS=FS=","}
      $2>a[$1]["max"]{a[$1]["max"]=$2}
      $2<a[$1]["min"] || !a[$1]["min"] {a[$1]["min"]=$2}
      END {for (i in a){
             print i, a[i]["min"],a[i]["max"]
             }
      }' file

示例here。

这里是另一种选择

$ join -t, <(sort file){,} | sort -k1,1 -k2n -k3nr | rev | uniq -2 | rev

使用sort、sed和awk的另一个答案

sort -k1,1 -k2n input.txt | sed -r ':a;N;$!ba;:b;s/(Task[0-9]+, )([0-9 ,]+)\n?\1([0-9]+)/\1\2, \3/g;tb;' | awk 'BEGIN{FS=OFS=", ";}{print $1, $2, $NF}'

仅使用sort和sed的替代解决方案

sort -k1,1 -k2n input.txt | sed -r ':a;N;$!ba;:b;s/(Task[0-9]+, )([0-9 ,]+)\n?\1([0-9]+)/\1\2, \3/g;tb;' | sed -r -e 's/^([^ ]+)\s([^ ]+)\s.*\s([^ ]+)/\1 \2 \3/' -e 's/^([^ ]+)\s([^ ]+)$/\1 \2, \2/'

你会得到

Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116

sort第一列和第二列，然后awk它。这个解决方案的好处（awk部分）是它不会将数据存储在内存中并最终转储出来，而是在找到新的$1后输出数据。这里：

$ sort -t, -k1 foo -k2n | \                     # sort 
awk '!($1 in min)    {min[$1]=$2}               # first of each is always min (and max)
      ($1 in min)    {max[$1]=$2}               # every current one is always max
       $1!=p && NR>1 {print p, min[p], max[p]}  # if $1 differs from previous, print previous
                     {p=$1}                     # p is current for next round
       END           {print p, min[p], max[p]}' # dump buffer
Task1, 1000 2100
Task2, 1100 3421
Task3, 1200 3300
Task4, 1299 1299
Task5, 1216 2216
Task6, 2416 2416
Task7, 1116 1116

这主要是bash，如果你对此有一些问题，我可以用其他东西代替awk命令...（例如colrm，如果时间总是在同一列中开始）。

# Keep a list of already processed task names
already_processed=""

# Use read to read only the first column from the data file
while IFS=',' read -ra task; do
  # If the task has already been processed, skip it and go to the next line
  if echo "$already_processed" | grep $task > /dev/null; then
    continue
  else
    # Select all the task with the same name from the data file, take the 
    #+second column and sort it to find the max and the minimum.
    MIN=`grep $task $1 | awk -F',' '{print $2}' | sort -n | head -1`
    MAX=`grep $task $1 | awk -F',' '{print $2}' | sort -n | tail -1`
    # Add the task to the "already_processed" tasks (to be sure each task will 
    #+appear only once in the output
    already_processed="$already_processed:$task"
    # Print the output in the wanted format.
    echo "${task}, ${MIN}, ${MAX}"
  fi

done < $1

只要确保数据文件以空行结尾即可。
示例：

bash <name_of_script_file> <name_of_data_file> | sort    
Task1, 1000, 2100
Task2, 1100, 3421
Task3, 1200, 3300
Task4, 1299, 1299
Task5, 1216, 2216
Task6, 2416, 2416
Task7, 1116, 1116