我试图替换文件中的现有数据值。该文件用于通过FEM代码计算运动材料硬化。每个元素的文本都是这样的。我有几个要修改的元素：
13499 8 10 1 1 0 0
1.500360e+08 1.451964e+07 3.728601e+08 -49529.937500 -1.930966e+07
4.499323e+06 4.427284e-03 2.107339e+06 -5.031084e+06 2.923746e+06
15250.131836 -1.833509e+06 1.992212e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.377865e+08 2.413883e+07 3.754904e+08 7.427730e+05 -2.500246e+07
2.964798e+06 4.422367e-03 1.855410e+06 -4.713138e+06 2.857730e+06
13224.164062 -2.054056e+06 1.624522e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.373486e+08 2.433924e+07 3.757281e+08 -4.595526e+06 -2.536434e+07
1.499050e+06 4.435852e-03 1.867437e+06 -4.746126e+06 2.878688e+06
-1.195839e+05 -2.081042e+06 1.472030e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.495412e+08 1.476622e+07 3.731083e+08 -5.636892e+06 -1.967031e+07
3.060086e+06 4.440636e-03 2.109059e+06 -5.047602e+06 2.938542e+06
-1.202835e+05 -1.849329e+06 1.847072e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.586573e+08 1.004740e+07 3.687111e+08 -6.710480e+05 -2.547207e+07
6.056666e+06 3.514634e-03 1.647607e+06 -3.689882e+06 2.042273e+06
417.019592 -1.201436e+06 1.671364e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.461364e+08 2.007670e+07 3.712027e+08 66494.984375 -3.248405e+07
4.652704e+06 3.476992e-03 1.358743e+06 -3.303262e+06 1.944519e+06
1658.200684 -1.316609e+06 1.339788e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.457837e+08 2.024077e+07 3.713914e+08 -5.707908e+06 -3.274767e+07
3.163598e+06 3.490632e-03 1.395223e+06 -3.361252e+06 1.966029e+06
-1.027537e+05 -1.351428e+06 1.227283e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
1.581765e+08 1.031220e+07 3.689273e+08 -6.650824e+06 -2.571731e+07
4.596290e+06 3.530744e-03 1.661256e+06 -3.721140e+06 2.059885e+06
-1.082121e+05 -1.219397e+06 1.555895e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
我想通过使用一些智能命令，脚本等将所有元素的文本更改为此文件。我使用的是Linux，但如果推荐使用Windows 7或Mac OS X也可以使用。我知道FORTRAN，Scilab和Octave的基础知识，但欢迎在这些环境之外的解决方案。
13499 8 10 1 1 0 0
0.000 0.000 0.000 0.000 0.000
0.000 0.000 2.107339e+06 -5.031084e+06 2.923746e+06
15250.131836 -1.833509e+06 1.992212e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.855410e+06 -4.713138e+06 2.857730e+06
13224.164062 -2.054056e+06 1.624522e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.867437e+06 -4.746126e+06 2.878688e+06
-1.195839e+05 -2.081042e+06 1.472030e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 2.109059e+06 -5.047602e+06 2.938542e+06
-1.202835e+05 -1.849329e+06 1.847072e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.647607e+06 -3.689882e+06 2.042273e+06
417.019592 -1.201436e+06 1.671364e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.358743e+06 -3.303262e+06 1.944519e+06
1658.200684 -1.316609e+06 1.339788e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.395223e+06 -3.361252e+06 1.966029e+06
-1.027537e+05 -1.351428e+06 1.227283e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000
0.000 0.000 1.661256e+06 -3.721140e+06 2.059885e+06
-1.082121e+05 -1.219397e+06 1.555895e+05 20.000000 0.000
1.000000 1.009244e-10 0.000 0.000 0.000
请告知。
答案 0 :(得分：2)
这是MATLAB / Octave中的解决方案。首先，我们将使用textread来读取实际数据。我看到你的文字数据不均匀。第一行有7个元素，而下一行有5个元素。 textread将对此进行zeropad并将数据作为单个矩阵读入。因此，第6列和第7列将具有除第一行之外的零。然后我提取出数字数据，我们只是跳过第一行，只抓取前5列。
在此之后，我循环遍历行，将第一行设置为零，将第一行和第二行设置为0的第二行，然后在此之后跳过接下来的两行。我重复这个过程，直到我们到达文本文件的末尾。我将第一列和这个新的数字数据保存到文本文件中。我将使用dlmwrite来促进这一写作。
因此，假设输入数据在testData.txt中，我将新文件写入testOut.txt，并遵循我们刚才谈到的上述规则。就这样：
%// Read in data and extract numeric data only
testData = textread('testData.txt');
numericData = testData(2:end,1:5);
count = 1;
%// From the first row until the last
while(count <= size(numericData,1))
numericData(count,:) = 0; %// Set first row to zero
%// Ensure we don't go out of bounds
if (count + 1 > size(numericData,1))
break;
end
numericData(count+1,1:2) = 0; %// Set first two elements of next row to zero
count = count + 4; %// Skip over two rows
end
%// Save the file. Write the first row first
dlmwrite('testOut.txt', testData(1,:), '\t');
%// Append on top of the file the modified numeric data
dlmwrite('testOut.txt', numericData, '-append', 'delimiter', '\t', 'precision', '%.6e');
我在这个文本文件中得到的输出是：
13499 8 10 1 1 0 0
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 2.107339e+06 -5.031084e+06 2.923746e+06
1.525013e+04 -1.833509e+06 1.992212e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.855410e+06 -4.713138e+06 2.857730e+06
1.322416e+04 -2.054056e+06 1.624522e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.867437e+06 -4.746126e+06 2.878688e+06
-1.195839e+05 -2.081042e+06 1.472030e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 2.109059e+06 -5.047602e+06 2.938542e+06
-1.202835e+05 -1.849329e+06 1.847072e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.647607e+06 -3.689882e+06 2.042273e+06
4.170196e+02 -1.201436e+06 1.671364e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.358743e+06 -3.303262e+06 1.944519e+06
1.658201e+03 -1.316609e+06 1.339788e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.395223e+06 -3.361252e+06 1.966029e+06
-1.027537e+05 -1.351428e+06 1.227283e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 1.661256e+06 -3.721140e+06 2.059885e+06
-1.082121e+05 -1.219397e+06 1.555895e+05 2.000000e+01 0.000000e+00
1.000000e+00 1.009244e-10 0.000000e+00 0.000000e+00 0.000000e+00
请注意，这是制表符分隔文件。祝你好运！
答案 1 :(得分：0)
我的问题的一个成功解决方案是由Andreas de Blanche http://www.hv.se/en/research/researchers/alphabetically/andreas-de-blanche提供的以下脚本：
#!/usr/bin/perl
use strict;
my $start = time;
my $inputName = $ARGV[0];
if($inputName)
{
print "\n Converting file: $inputName\n";
print " Output file: $inputName.out.\n\n";
}
else
{
print "\n Please enter input filename on commandline\n";
print " ./script.pl <input.txt>\n";
print " Output file: <input.out>.\n\n";
exit;
}
open INFILE, "<$inputName" or die $!;
open OUTFILE, ">$inputName.out" or die $!;
my $cardRow=33; #Each card is 33 rows long.
my $count=0;
my $row=0;
while (<INFILE>)
{
# print "$row: $_";
if($_=="")
{
}
elsif($cardRow==33)
{
#print Reset counter $count
$count=-1;
$cardRow=0;
#Just print line to outfile
print OUTFILE "$_";
}
elsif($count==0)
{
#These rows seems to always be identical
print OUTFILE " 0.000000000E+00 0.000000000E+00 0.000000000E+00 0.000000000E+00 0.000000000E+00\n";
}
elsif($count==1)
{
my $fragment = substr $_, 32;
print OUTFILE " 0.000000000E+00 0.000000000E+00$fragment";
}
elsif($count==2 || $count==3)
{
#These rows should not be changed
print OUTFILE "$_";
}
else
{
print "ERROR: This wrong sized subset!\n";
}
$count++;
if($count==4)
{
#There are 4 rows in each subset (0-3)
$count=0;
}
$cardRow++;
$row++;
}
close INFILE;
close OUTFILE;
my $duration = time - $start;
print " Execution time: $duration s\n";