Question

我有一个由空格分隔的3个数字的大单元格数组（下面是150万的前5行）。

LocationCell = 
'3926.611     -1534.095        26.324'
'4122.978       882.279       -67.495'
'4078.042      1072.946        60.384'
'4047.521     -1182.700        10.520'
'4188.222      -468.615       -57.303'

数组太大，需要几个小时才能循环并提取数字。

for n = 1: (1.5million)
    LocationNumbers(n,:) = textscan( LocationCell{n}, '%f %f %f ');
end

有没有办法可以在没有循环的情况下做到这一点？

渴望出局：

LocationNumbers = 
3926.611, -1534.095,  26.324;
4122.978,   882.279, -67.495;
4078.042,  1072.946,  60.384;
4047.521, -1182.700,  10.520;
4188.222,  -468.615, -57.303

Answer 1

可能需要几个小时，因为你不是preallocating，迫使MATLAB不断搜索新的连续内存块而不是从一开始就分配正确的大小块，从而导致显着的性能损失。预分配和sscanf输出正确的数据类（textscan输出一个单元格数组），大大缩短了时间：

LocationCell = {'3926.611     -1534.095        26.324';
                '4122.978       882.279       -67.495';
                '4078.042      1072.946        60.384';
                '4047.521     -1182.700        10.520';
                '4188.222      -468.615       -57.303'};

ncells = numel(LocationCell);
LocationNumbers = zeros(ncells, 3);
for n = 1:(ncells)
    LocationNumbers(n,:) = sscanf(LocationCell{n}, '%f %f %f');
end

在R2018a中提供以下内容（下面的完整时序代码）：

Timing Results
n cells:  125000
================
Original: 6.638
regex:    3.840
strsplit: 11.957
sscanf:   0.958

LocationCell = repmat({'3926.611     -1534.095        26.324'; ...
                       '4122.978       882.279       -67.495'; ...
                       '4078.042      1072.946        60.384'; ...
                       '4047.521     -1182.700        10.520'; ...
                       '4188.222      -468.615       -57.303'}, ...
                      25000, 1);

t1 = timeit(@()thing1(LocationCell));
t2 = timeit(@()thing2(LocationCell));
t3 = timeit(@()thing3(LocationCell));
t4 = timeit(@()thing4(LocationCell));

fprintf(['Timing Results\n', ...
         'n cells:  %u\n', ...
         '================\n', ...
         'Original: %0.3f\n', ...
         'regex:    %0.3f\n', ...
         'strsplit: %0.3f\n', ...
         'sscanf:   %0.3f\n'], numel(LocationCell), t1, t2, t3, t4)

function out = thing1(in)
for n = 1: (numel(in))
    out(n,:) = textscan(in{n}, '%f %f %f ');
end
end

function out = thing2(in)
S = regexp(in,'(\-?\d+\.\d+)[ ]+(\-?\d+\.\d+)[ ]+(\-?\d+\.\d+)','tokens','once');
S = vertcat(S{:});
out = str2double(S);
end

function out = thing3(in)
S = cellfun(@(x)strsplit(x,' '), in, 'UniformOutput', false);
S = vertcat(S{:});
out = str2double(S);
end

function out = thing4(in)
ncells = numel(in);
out = zeros(ncells, 3);
for n = 1:(numel(in))
    out(n,:) = sscanf(in{n}, '%f %f %f ');
end
end

Answer 2

这应该可以解决问题：

C = {
  '3926.611     -1534.095        26.324';
  '4122.978       882.279       -67.495';
  '4078.042      1072.946        60.384';
  '4047.521     -1182.700        10.520';
  '4188.222      -468.615       -57.303'
};

% Split the cell elements using a regular expression...
S = regexp(C,'(\-?\d+\.\d+)[ ]+(\-?\d+\.\d+)[ ]+(\-?\d+\.\d+)','tokens','once');

% Flatten the result...
S = vertcat(S{:});

% Convert the cell matrix to double...
M = str2double(S)

或者，您可以继续进行传统拆分，而不是使用正则表达式：

C = {
  '3926.611     -1534.095        26.324';
  '4122.978       882.279       -67.495';
  '4078.042      1072.946        60.384';
  '4047.521     -1182.700        10.520';
  '4188.222      -468.615       -57.303'
};

% Split the cell elements...
S = cellfun(@(x)strsplit(x,' '),C,'UniformOutput',false);

% Flatten the result...
S = vertcat(S{:});

% Convert the cell matrix to double...
M = str2double(S)

Matlab：循环大量将单元格数组转换为数字

2 个答案: