我有两个不同的文件。部分文件是:
文件1:
ifile1.txt
21 10.3 70.32 09
32 11.3 71.43 10
33 15.4 75.00 00
54 17.3 68.03 12
95 19.2 65.02 21
99 20.1 80.10 11
and so on......
Where 1st column is the ID,
2nd column refers to x-axis,
3rd column refers to y-axis,
4th column refers to value
file2的:
ifile2.txt
10.10 70.12 10
10.11 73.33 08
11.05 72.40 00
11.30 69.13 15
12.00 64.02 27
12.05 79.20 25
13.10 80.32 10
13.11 75.43 06
14.05 74.00 02
14.20 69.03 15
16.40 65.02 13
16.55 68.10 14
and so on......
Where 1st column refers to x-axis,
2nd column refers to y-axis,
3rd column refers to value
我想制作另一个文件,其输出将是(对于上面的例子):
ofile.txt
21 10.10 70.12 10
32 11.05 71.40 10
33 14.05 74.00 02
54 16.55 68.10 14
95 16.55 68.10 14
99 16.55 68.10 14
Where 1st column is the ID (taken from ifile1.txt),
2nd and 3rd column refers to x-axis and y-axis (taken from ifile2.txt which are the nearest to those of ifile1.txt),
4th column refers to corresponding value (taken from ifile2.txt)
我无法为此编写shell脚本。但是我的算法是:
for each $1 in ifile1.txt,
for each $2 & $3 in ifile1.txt,
find a $2 & $3 in ifile2.txt such that
sqrt{[$2(ifile1.txt)-$2(ifile2.txt)]^2+[$3(ifile1.txt)-$3(ifile2.txt)]^2} is minimum for all $2 & $3 in ifile2.txt
find correcsponding $4 from ifile2.txt
write $1, $2, $3, $4 to ofile.txt
我非常感谢任何建议或帮助。
答案 0 :(得分:1)
这是awk中的一个简单解决方案。它在我的Linux上运行良好。 如果我的回答对你有用,请点击我答案左侧的灰色勾号。
#!/bin/bash
(($#!=2))&& { echo "Usage $0 1st_file 2nd_file"; exit 1; }
awk '
BEGIN {p=fx=0; fn=""; maxd=1.1e11;}
$0~"[^0-9. \t]" || NF!=4 && NF!=3 {next;} # skip no data lines
fn!=FILENAME {fx++; fn=FILENAME;} # fx: which file
fx==1 { if(NF!=3){printf("Change the series of two input files\n"); exit 1;}
x2[p]=$1; y2[p]=$2; d2[p++]=$3;next;} # save the columns of first file
fx==2 { mv=maxd; mp=0; # search minimal distance
for(i=0; i<p; i++){
dx=x2[i]-$2; dy=y2[i]-$3; dd=dx*dx+dy*dy;
if(dd<mv){mv=dd; mp=i;} # min value & min place
}
printf("%3d %6.2f %6.2f %3d\n", $1, x2[mp], y2[mp], d2[mp]);
}
' $2 $1 # first is the 2nd_file!
答案 1 :(得分:0)
我对这个问题的看法如下:
awk '(NR==FNR) { index[FNR]=$1; coordx[FNR]=$2; coordy[FNR]=$3; val[FNR=$4]; next }
{ for(i in index) {
d2=(coordx[i]-$1)*(coordx[i]-$1) + (coordy[i]-$2)*(coordy[i]-$2)
sol[i] = (i in sol ? (min_dist[i] > d2 ? $0 : sol[i]) : $0 )
min_dist[i] = (sol[i] == $0 ? d2 : min_dist[i])
}
END { for(i in index) { print index[i], sol[i] } }
' file1 file2