如何计算第一列中具有相同ID的数据的平均值（每列）？

Question 1

这是一个 Perl 解决方案。它并不期望对年份进行排序，但确实需要将整个文件（至少是每列的总和）读入内存。只要你的文件不是太大，它应该做你想要的：

#!/usr/bin/env perl

## Read the input file, line by line
while (<>){
    ## Print and skip the 1st line
    if($.==1){
    print; 
    next
    }; 
    ## Split the fileds into the array @F
    @F=split(/\s+/);

    ## $k (%k) is a hash of hashes. The first element of 
    ## $F ($F[0]) is the year, so $k{$year}{tot} will be 
    ## the nuber of lines for each year.
    $k{$F[0]}{tot}++;

    ## Go through each field add its value
    ## to $k{$year}{field number}
    for($i=1;$i<=$#F;$i++){
    $k{$F[0]}{$i}+=$F[$i];
    }
 }
## Now print the data for each year
foreach $y (sort keys(%k)){
    ## $y is the year
    print "$y "; 
    ## $k{$y}{$i} is the sum of the values for this column/year
    ## and $k{$y}{tot} is the number of lines for this year
    for($i=1;$i<=$#F;$i++){
    print " " . $k{$y}{$i}/$k{$y}{tot};

    }
    print "\n";
}

这是与（较长的）单行相同的脚本：

$ perl -ane '$.==1 && do {print; next}; $k{$F[0]}{tot}++;
           for($i=1;$i<=$#F;$i++){$k{$F[0]}{$i}+=$F[$i];}
           END{foreach $y (sort keys(%k)){
                    print "$y "; 
                    for($i=1;$i<=$#F;$i++){
                      print " " . $k{$y}{$i}/$k{$y}{tot};
                    } print "\n";} }' abraham 
year    1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54
1929  4.6525 0 4.7245 2.9415 1.05175 0.21 -0.06275 -0.19975 -0.3185 -0.42875 -0.533 -0.63225 -0.72775 -0.90775 -1.07525 -1.23075 -1.37525 -1.50925 -1.803 -2.043 -2.23225 -2.37475 -2.47525 -2.5375 -2.571 -2.51425 -2.40125 -2.262 -2.11925 -1.9885 -1.8735 -1.776 -1.69225 -1.62925 -1.5785 -1.4845 -1.39525 -1.311 -1.229 -1.151 -1.075 -1.002 -0.859 -0.721 -0.587 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.308 0.445 0.559
1930  1.25 0.0005 1.2375 0.65475 0.251 0.049 -0.052 -0.184 -0.30075 -0.409 -0.51125 -0.6085 -0.702 -0.8775 -1.041 -1.193 -1.33475 -1.46675 -1.757 -1.99475 -2.18425 -2.32875 -2.432 -2.499 -2.54225 -2.49625 -2.39325 -2.26075 -2.1225 -1.99275 -1.878 -1.7795 -1.6955 -1.6315 -1.58075 -1.48575 -1.39625 -1.31175 -1.23 -1.15175 -1.076 -1.002 -0.85925 -0.721 -0.587 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.308 0.445 0.55875
1931  -8.4025 0.057 -5.51075 -1.187 0.0095 0.001 -0.055 -0.1735 -0.2835 -0.38725 -0.4855 -0.57975 -0.67025 -0.84075 -0.9995 -1.148 -1.28625 -1.41575 -1.70075 -1.9365 -2.1255 -2.2715 -2.379 -2.45125 -2.50775 -2.47425 -2.38225 -2.25825 -2.125 -1.99725 -1.883 -1.7835 -1.69825 -1.63325 -1.58275 -1.487 -1.3975 -1.31225 -1.231 -1.152 -1.076 -1.003 -0.86 -0.722 -0.58725 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.307 0.444 0.558

Answer

这是一个 Perl 解决方案。它并不期望对年份进行排序，但确实需要将整个文件（至少是每列的总和）读入内存。只要你的文件不是太大，它应该做你想要的：

#!/usr/bin/env perl

## Read the input file, line by line
while (<>){
    ## Print and skip the 1st line
    if($.==1){
    print; 
    next
    }; 
    ## Split the fileds into the array @F
    @F=split(/\s+/);

    ## $k (%k) is a hash of hashes. The first element of 
    ## $F ($F[0]) is the year, so $k{$year}{tot} will be 
    ## the nuber of lines for each year.
    $k{$F[0]}{tot}++;

    ## Go through each field add its value
    ## to $k{$year}{field number}
    for($i=1;$i<=$#F;$i++){
    $k{$F[0]}{$i}+=$F[$i];
    }
 }
## Now print the data for each year
foreach $y (sort keys(%k)){
    ## $y is the year
    print "$y "; 
    ## $k{$y}{$i} is the sum of the values for this column/year
    ## and $k{$y}{tot} is the number of lines for this year
    for($i=1;$i<=$#F;$i++){
    print " " . $k{$y}{$i}/$k{$y}{tot};

    }
    print "\n";
}

这是与（较长的）单行相同的脚本：

$ perl -ane '$.==1 && do {print; next}; $k{$F[0]}{tot}++;
           for($i=1;$i<=$#F;$i++){$k{$F[0]}{$i}+=$F[$i];}
           END{foreach $y (sort keys(%k)){
                    print "$y "; 
                    for($i=1;$i<=$#F;$i++){
                      print " " . $k{$y}{$i}/$k{$y}{tot};
                    } print "\n";} }' abraham 
year    1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54
1929  4.6525 0 4.7245 2.9415 1.05175 0.21 -0.06275 -0.19975 -0.3185 -0.42875 -0.533 -0.63225 -0.72775 -0.90775 -1.07525 -1.23075 -1.37525 -1.50925 -1.803 -2.043 -2.23225 -2.37475 -2.47525 -2.5375 -2.571 -2.51425 -2.40125 -2.262 -2.11925 -1.9885 -1.8735 -1.776 -1.69225 -1.62925 -1.5785 -1.4845 -1.39525 -1.311 -1.229 -1.151 -1.075 -1.002 -0.859 -0.721 -0.587 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.308 0.445 0.559
1930  1.25 0.0005 1.2375 0.65475 0.251 0.049 -0.052 -0.184 -0.30075 -0.409 -0.51125 -0.6085 -0.702 -0.8775 -1.041 -1.193 -1.33475 -1.46675 -1.757 -1.99475 -2.18425 -2.32875 -2.432 -2.499 -2.54225 -2.49625 -2.39325 -2.26075 -2.1225 -1.99275 -1.878 -1.7795 -1.6955 -1.6315 -1.58075 -1.48575 -1.39625 -1.31175 -1.23 -1.15175 -1.076 -1.002 -0.85925 -0.721 -0.587 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.308 0.445 0.55875
1931  -8.4025 0.057 -5.51075 -1.187 0.0095 0.001 -0.055 -0.1735 -0.2835 -0.38725 -0.4855 -0.57975 -0.67025 -0.84075 -0.9995 -1.148 -1.28625 -1.41575 -1.70075 -1.9365 -2.1255 -2.2715 -2.379 -2.45125 -2.50775 -2.47425 -2.38225 -2.25825 -2.125 -1.99725 -1.883 -1.7835 -1.69825 -1.63325 -1.58275 -1.487 -1.3975 -1.31225 -1.231 -1.152 -1.076 -1.003 -0.86 -0.722 -0.58725 -0.456 -0.328 -0.171 -0.017 0.068 0.151 0.307 0.444 0.558

Question 2

您可以使用数组来存储每列值的总和，当 $1 发生变化时，从那里获取平均值。

# print header
NR==1 {
    print
    next
}

# print average of each column per year
#  then, reset columns sums and number of lines
function print_sum() {
    printf prev
    # needs GNU awk, for length of array
    for (i=2; i < length(sum) + 2; i++) {
            printf FS sum[i]/nlines
            sum[i] = 0
    }
    printf ORS
    nlines = 0
}

# print average when $1 changes, but not the first time
# also, on end of script
NR>2 && prev!=$1 { print_sum() }
END              { print_sum() }

# for every line with the same $1, sum column values, increment number of lines
{
    prev=$1;
    nlines++
    for (i=2; i <= NF; i++) {
            sum[i]+=$i
    }
}

Answer

您可以使用数组来存储每列值的总和，当 $1 发生变化时，从那里获取平均值。

# print header
NR==1 {
    print
    next
}

# print average of each column per year
#  then, reset columns sums and number of lines
function print_sum() {
    printf prev
    # needs GNU awk, for length of array
    for (i=2; i < length(sum) + 2; i++) {
            printf FS sum[i]/nlines
            sum[i] = 0
    }
    printf ORS
    nlines = 0
}

# print average when $1 changes, but not the first time
# also, on end of script
NR>2 && prev!=$1 { print_sum() }
END              { print_sum() }

# for every line with the same $1, sum column values, increment number of lines
{
    prev=$1;
    nlines++
    for (i=2; i <= NF; i++) {
            sum[i]+=$i
    }
}

Question 3

和gnu datamash:

{ head -n 1; datamash -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

其中54是列数。这假设您的字段是制表符分隔的。如果它们是空间分隔的运行

{ head -n 1; datamash -t ' ' -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

或者如果它们由多个空格（空格/制表符）分隔：

{ head -n 1; datamash -W -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

请注意，输出将按第一列排序。

Answer

和gnu datamash:

{ head -n 1; datamash -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

其中54是列数。这假设您的字段是制表符分隔的。如果它们是空间分隔的运行

{ head -n 1; datamash -t ' ' -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

或者如果它们由多个空格（空格/制表符）分隔：

{ head -n 1; datamash -W -s -g 1 $(printf 'mean %s ' {2..54}); } <infile

请注意，输出将按第一列排序。

如何计算第一列中具有相同ID的数据的平均值（每列）？

答案1

答案2

答案3

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