使用awk将列转换为矩阵格式 [英] Convert column to matrix format using awk

问题描述

我有一个列格式的网格数据文件，如下:

I have a gridded data file in column format as:

ifile.txt
x     y     value
20.5  20.5  -4.1
21.5  20.5  -6.2
22.5  20.5   0.0
20.5  21.5   1.2
21.5  21.5   4.3
22.5  21.5   6.0
20.5  22.5   7.0
21.5  22.5  10.4
22.5  22.5  16.7

我想将其转换为矩阵格式，如下:

I would like to convert it to matrix format as:

ofile.txt
     20.5  21.5 22.5
20.5 -4.1   1.2  7.0
21.5 -6.2   4.3 10.4
22.5  0.0   6.0 16.7

顶部的20.5 21.5 22.5表示y，侧面的值表示x，内部的值表示对应的网格值.

Where top 20.5 21.5 22.5 indicate y and side values indicate x and the inside values indicate the corresponding grid values.

我在这里找到了类似的问题将3列文件转换为矩阵格式，但该脚本在我的情况下不起作用.

I found a similar question here Convert a 3 column file to matrix format but the script is not working in my case.

脚本是

awk '{ h[$1,$2] = h[$2,$1] = $3 }
    END {
      for(i=1; i<=$1; i++) {
        for(j=1; j<=$2; j++)
          printf h[i,j] OFS
        printf "\n"
      }
    }' ifile

推荐答案

以下awk脚本句柄:

• 任何大小的矩阵
• 行索引和列索引之间没有关系，因此可以分别跟踪它们.
• 如果未出现某个行列索引，则该值将默认为零.

这是通过以下方式完成的:

This is done in this way:

awk '
BEGIN{PROCINFO["sorted_in"] = "@ind_num_asc"}
(NR==1){next}
{row[$1]=1;col[$2]=1;val[$1" "$2]=$3}
END { printf "%8s",""; for (j in col) { printf "%8.3f",j }; printf "\n"
      for (i in row) {
        printf "%8.3f",i; for (j in col) { printf "%8.3f",val[i" "j] }; printf "\n"
      }
    }' <file>

它是如何工作的:

• PROCINFO["sorted_in"] = "@ind_num_asc"，指出所有数组均按索引按数字排序.
• (NR==1){next}:跳过第一行
• {row[$1]=1;col[$2]=1;val[$1" "$2]=$3}，通过存储行和列索引以及随附的值来处理行.
• end语句完成所有打印.

• PROCINFO["sorted_in"] = "@ind_num_asc", states that all arrays are sorted numerically by index.
• (NR==1){next} : skip the first line
• {row[$1]=1;col[$2]=1;val[$1" "$2]=$3}, process the line by storing the row and column index and accompanying value.
• The end statement does all the printing.

这将输出:

          20.500  21.500  22.500
  20.500  -4.100   1.200   7.000
  21.500  -6.200   4.300  10.400
  22.500   0.000   6.000  16.700

注意:PROCINFO的使用是gawk功能.

但是，如果您做几个假设，则可以做得更短:

However, if you make a couple of assumptions, you can do it much shorter:

• 该文件包含所有可能的条目，没有丢失的值
• 您不希望打印出行和列的索引:
• 索引按列主要顺序
进行排序>

• the file contains all possible entries, no missing values
• you do not want the indices of the rows and columns printed out:
• the indices are sorted in column-major-order

您可以使用以下简短版本:

The you can use the following short versions:

sort -g <file> | awk '($1+0!=$1){next}
                      ($1!=o)&&(NR!=1){printf "\n"}
                      {printf "%8.3f",$3; o=$1 }'

输出

  -4.100   1.200   7.000
  -6.200   4.300  10.400
   0.000   6.000  16.700

或换位:

awk '(NR==1){next}
     ($2!=o)&&(NR!=2){printf "\n"}
     {printf "%8.3f",$3; o=$2 }' <file>

此输出

  -4.100  -6.200   0.000
   1.200   4.300   6.000
   7.000  10.400  16.700

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