有没有办法在单个命令中获取数字列表的最小值，最大值，中位数和平均值？

#1 楼

#! /usr/bin/env Rscript
d<-scan("stdin", quiet=TRUE)
cat(min(d), max(d), median(d), mean(d), sep="\n")

$ cat datafile
1
2
4
$ ./mmmm.r < datafile
1
4
2
2.333333

$ cat datafile2
1.1
2.2
4.4
$ ./mmmm.r < datafile2
1.1
4.4
2.2
2.566667

$ Rscript -e 'd<-scan("stdin", quiet=TRUE)' \
          -e 'cat(min(d), max(d), median(d), mean(d), sep="\n")' < datafile
1
4
2
2.333333

> summary(c(1,2,4))
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  1.000   1.500   2.000   2.333   3.000   4.000 

#2 楼

$ printf '%s\n' 1 2 4 | datamash max 1 min 1 mean 1 median 1
4   1   2.3333333333333 2

#3 楼

#!/bin/sh
sort -n | awk '
  BEGIN {
    c = 0;
    sum = 0;
  }
   ~ /^(\-)?[0-9]*(\.[0-9]*)?$/ {
    a[c++] = ;
    sum += ;
  }
  END {
    ave = sum / c;
    if( (c % 2) == 1 ) {
      median = a[ int(c/2) ];
    } else {
      median = ( a[c/2] + a[c/2-1] ) / 2;
    }
    OFS="\t";
    print sum, c, ave, median, a[0], a[c-1];
  }
'

#4 楼

 jq -s min
awk 'NR==1||jq -s max
awk 'NR==1||jq -s 'sort|if length%2==1 then.[length/2|floor]else[.[length/2-1,length/2]]|add/2 end'
sort -n|awk '{a[NR]=jq -s add/length
awk '{x+=$ seq 100|jq -s '{minimum:min,maximum:max,average:(add/length),median:(sort|if length%2==1 then.[length/2|floor]else[.[length/2-1,length/2]]|add/2 end)}'
{
  "minimum": 1,
  "maximum": 100,
  "average": 51.5,
  "median": 51.5
}
}END{print x/NR}'
}END{print(NR%2==1)?a[int(NR/2)+1]:(a[NR/2]+a[NR/2+1])/2}'
>x{x=jq}END{print x}'
<x{x=-s}END{print x}'
 

 --slurp 

 q4312079q 

 q4312079q 

 q4312079q 

#5 楼

% echo -e '6\n2\n4\n3\n1' | awk 'NR == 1 { max=; min=; sum=0 }
   { if (>max) max=; if (<min) min=; sum+=;}
   END {printf "Min: %d\tMax: %d\tAverage: %f\n", min, max, sum/NR}'
Min: 1  Max: 6  Average: 3,200000

% echo -e '6\n2\n4\n3\n1' | sort -n | awk '{arr[NR]=}
   END { if (NR%2==1) print arr[(NR+1)/2]; else print (arr[NR/2]+arr[NR/2+1])/2}' 
3

#6 楼

cat file.txt | py --ji -l 'min(l), max(l), numpy.median(l), numpy.mean(l)'

#7 楼

cat numbers.txt \
| perl -M'List::Util qw(sum max min)' -MPOSIX -0777 -a -ne 'printf "%-7s : %d\n"x4, "Min", min(@F), "Max", max(@F), "Average", sum(@F)/@F,  "Median", sum( (sort {$a<=>$b} @F)[ int( $#F/2 ), ceil( $#F/2 ) ] )/2;'

#!/usr/bin/perl

use List::Util qw(sum max min);
use POSIX;

@F=<>;

printf "%-7s : %d\n" x 4,
    "Min", min(@F),
    "Max", max(@F),
    "Average", sum(@F)/@F,
    "Median", sum( (sort {$a<=>$b} @F)[ int( $#F/2 ), ceil( $#F/2 ) ] )/2;

#8 楼

nums=$(<file.txt); 
list=(`for n in $nums; do printf "%015.06f\n" $n; done | sort -n`); 
echo min ${list[0]}; 
echo max ${list[${#list[*]}-1]}; 
echo median ${list[${#list[*]}/2]};

#9 楼

octave -q --eval 'A=1:10;
  printf ("# %f\t%f\t%f\t%f\n", min(A), max(A), median(A), mean(A));'  
# 1.000000        10.000000       5.500000        5.500000

arr=($(sort -n "LIST" |tee >(numaverage 2>/dev/null >stats.avg) ))
cnt=${#arr[@]}; ((cnt==0)) && { echo -e "0\t0\t0\t0\t0"; exit; }
mid=$((cnt/2)); 
if [[ ${cnt#${cnt%?}} == [02468] ]] 
   then med=$( echo -n "${arr[mid-1]}" |numprocess /+${arr[mid]},%2/ )
   else med=${arr[mid]}; 
fi     #  count   min       max           median        average
echo -ne "$cnt\t${arr[0]}\t${arr[cnt-1]}\t$med\t"; cat stats.avg 

#10 楼

r summary file.txt
r -e 'min(d); max(d); median(d); mean(d)' file.txt

#11 楼

#!/usr/bin/env Rscript
a <- scan(file("stdin"), c(0), quiet=TRUE);
cat(min(a), max(a), mean(a), median(a), "\n");

#12 楼

$ echo "1 2 3 4 5 6 7 8 9" | num max
9
$ echo "1 2 3 4 5 6 7 8 9" | num min max median mean
..and so on

#13 楼

sort -n | awk '{a[i++]=q4312078q;s+=q4312078q}END{print a[0],a[i-1],(a[int(i/2)]+a[int((i-1)/2)])/2,s/i}'

#14 楼

6.0
4.2
8.3
9.5
1.7

5
1.7
4.2
6.0
8.3
9.5

FILENAME="Salaries.csv"

(awk 'BEGIN {c=0}  ~ /^[-0-9]*(\.[0-9]*)?$/ {c=c+1;} END {print c;}' "$FILENAME"; \
        sort -n "$FILENAME") | awk '
  BEGIN {
    c = 0
    sum = 0
    med1_loc = 0
    med2_loc = 0
    med1_val = 0
    med2_val = 0
    min = 0
    max = 0
  }

  NR==1 {
    LINES = 
    # We check whether numlines is even or odd so that we keep only
    # the locations in the array where the median might be.
    if (LINES%2==0) {med1_loc = LINES/2-1; med2_loc = med1_loc+1;}
    if (LINES%2!=0) {med1_loc = med2_loc = (LINES-1)/2;}
  }

   ~ /^[-0-9]*(\.[0-9]*)?$/  &&  NR!=1 {
    # setting min value
    if (c==0) {min = ;}
    # middle two values in array
    if (c==med1_loc) {med1_val = ;}
    if (c==med2_loc) {med2_val = ;}
    c++
    sum += 
    max = 
  }
  END {
    ave = sum / c
    median = (med1_val + med2_val ) / 2
    print "sum:" sum
    print "count:" c
    print "mean:" ave
    print "median:" median
    print "min:" min
    print "max:" max
  }
'

#15 楼

$ printf '%s\n' 1 2 4 |
   perl -MList::Util=min,max -MStatistics::Basic=mean,median -w -le '
     chomp(@l = <>); print for min(@l), max(@l), mean(@l), median(@l)'
1
4
2.33
2

#16 楼

R -q -e 'summary(as.numeric(read.table("your_single_col_file")[,1]))'

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  550.4   628.3   733.1   706.5   778.4   832.9

#17 楼

cat data |  python3 -c "import fileinput as FI,statistics as STAT; i = [int(l) for l in FI.input()]; print('min:', min(i), ' max: ', max(i), ' avg: ', STAT.mean(i), ' median: ', STAT.median(i))"

#18 楼

$max = 0;
while (<>) {
 $sum = $sum + $_;
 $max = $_ if ($_ > $max);
 $count++;
}
$avg=$sum/$count;
print "$count numbers total=$sum max=$max mean=$avg\n";

#19 楼

function median()
{
    declare -a nums=($(cat))
    printf '%s\n' "${nums[@]}" | sort -n | tail -n $((${#nums[@]} / 2 + 1)) | head -n 1
}  

#20 楼

{
  "min": 1,
  "max": 5,
  "sum": 15,
  "count": 5,
  "avg": 3,
  "median": 3
}