我有这种格式的文本文件:
Food
Fruits1 [heading]
Apple [value]
Mango [value]
Orange [value]
Veg1 [heading]
Potato [value]
Lettuce [value]
我想将此格式加载到 Octave 中作为此格式的矩阵:
----------------------------------------------------------------------
Item | Fruits1 | Apple | Mango | Orange | Veg1 | Potato | Lettuce
----------------------------------------------------------------------
| | | | | | |
----------------------------------------------------------------------
因此,我需要一个size = 2x(n + m + 1)的矩阵;其中n = [标题]的数量,m = [值]的数量。
如何使用 fgetl 从文本文件中读取每一行并存储到满足上述条件的矩阵中?有更好的想法吗?
谢谢!
编辑:代码: -
fid = fopen('food.txt','r');
num = 1;
if (fid < 0)
printf('Error:could not open file\n')
else
while ~feof(fid),
line = fgetl(fid);
arr=[line;];
num=num+1;
end;
fclose(fid)
end;
答案 0 :(得分:0)
无法使用矩阵,因为您的字符串长度不同。它只能将它放入一个细胞中。但我会推荐一个结构化的清单。
fid = fopen('list.txt');
while 1
tmp = fgetl(fid);
if ~ischar(tmp)
% end of file
break
end
if strcmp(deblank(tmp), 'Food')
# this can't be empty
listObj.('item') = 'Food';
else
tmp = cell2mat(regexp(deblank(tmp),'(\w+)','tokens'));
if strcmp(tmp{1,2}, 'heading')
head = tmp{1,1};
else
listObj.(head).(tmp{1,1}) = str2double(tmp{1,2});
end
end
end
您可以更好地访问它。
>> stackoverflow
>> listObj
listObj =
scalar structure containing the fields:
item = Food
Fruits1 =
scalar structure containing the fields:
Apple = 3
Mango = 4
Orange = 1
Veg1 =
scalar structure containing the fields:
Potato = 8
Lettuce = 0
>> listObj.Fruits1.Apple
ans = 3
>> listObj.Veg1.Lettuce
ans = 0
>>
但请注意您的输入文件,必须严格格式化。我的示例文件看起来像这样
Food
Fruits1 [heading]
Apple 3
Mango 4
Orange 1
Veg1 [heading]
Potato 8
Lettuce 0