Question

我试图从字符串中读取未知数量的数值。

例如：

line = 'bla bal bli : 3 5 12 15 266'

我试过sscanf，但它需要我知道字符串中有多少数字。

Answer 1

您只需使用isstrprop。

line = 'bla bal bli : 3 5 12 15 266'
line(isstrprop(line, 'digit'))

您从字符串中提取数字：

line =

bla bal bli : 3 5 12 15 266


ans =

351215266

Answer 2

如果要从字符串中提取整数，可以使用正则表达式：

将数字作为字符串的单元格数组：

>> line = 'bla bal bli : 3 5 12 15 266';
>> regexp(line,'(?<!\d)(\d)+(?!\d)','match')
ans = 
    '3'    '5'    '12'    '15'    '266'

要将数字作为数字向量：

>> str2double(regexp(line,'(?<!\d)(\d)+(?!\d)','match'))
ans =
     3     5    12    15   266

Answer 3

如果你的字符串在字母和数字之间总是有一个像冒号（':'）这样的分隔符，你可以使用strsplit只用数字来提取字符串的一部分，这将是比使用正则表达式更快：

line = 'bla bal bli : 3 5 12 15 266';
numstr = strsplit(line,':');

然后您可以使用sscanf将数字提取为双精度数组：

nums = sscanf(numstr{2},'%u') % Assuming digits are always unsigned, if not use %d

如果您使用的是R2012b或更早版本，则不会有strsplit。有关替代方案，请参阅this answer。

Answer 4

如果有可能是负数或十进制数，正则表达式可以这样更新：

 1  function Dijkstra(Graph, source):
 2
 3      create vertex set Q
 4
 5      for each vertex v in Graph:             // Initialization
 6          dist[v] ← INFINITY                  // Unknown distance from source to v
 7          prev[v] ← UNDEFINED                 // Previous node in optimal path from source
 8          add v to Q                          // All nodes initially in Q (unvisited nodes)
 9
10      dist[source] ← 0                        // Distance from source to source
11      
12      while Q is not empty:
13          u ← vertex in Q with min dist[u]    // Node with the least distance will be selected first
14          remove u from Q 
15          
16          for each neighbor v of u:           // where v is still in Q.
17              alt ← dist[u] + length(u, v)
18              if alt < dist[v]:               // A shorter path to v has been found
19                  dist[v] ← alt 
20                  prev[v] ← u 
21
22      return dist[], prev[]

示例：

str2double(regexp(line,'(-?\d+\.?\d?)','match'))

ans =

3 5 12 15 266

str2double(regexp('bla bal bli : 3 5 12 15 266','(-?\d+\.?\d?)','match'))

ans =

3.0000 5.0000 12.0000 15.4000 -266.0000

Answer 5

从16b开始，您可以使用字符串类

line    = "bla bal bli : 3 5 12 15 266";
numbers = extractAfter(line,': ');
numbers = split(numbers);
numbers = double(numbers);

这似乎是最快的选择。

>> profFunc
Elapsed time is 0.088259 seconds.
Elapsed time is 1.558718 seconds.
Elapsed time is 1.482743 seconds.

function profFunc

    n = 1E4;

    tic;
        for i = 1:n
            line = "bla bal bli : 3 5 12 15 266";
            numbers = extractAfter(line,': ');
            numbers = split(numbers);
            numbers = double(numbers);
        end
    toc

    tic;
        for i = 1:n
            line = 'bla bal bli : 3 5 12 15 266';
            numbers = str2double(regexp(line,'(?<!\d)(\d)+(?!\d)','match'));
        end
    toc

    tic;
        for i = 1:n
            line = 'bla bal bli : 3 5 12 15 266';
            numstr = strsplit(line,':');
            nums = sscanf(numstr{2},'%u');
        end
    toc
end

MATLAB从字符串中读取所有数字

5 个答案: