blocks|key|724368|text|是的，检查这个：http://www.obihiro.ac.jp/~suzukim/masuda/octave/html3/octave_36.html#SEC75|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|724369|a+=+["hello";+"world"];
c+=+cellstr+(a)
+++++⇒+c+=
+++++++++{
+++++++++++[1,1]+=+hello
+++++++++++[2,1]+=+world
+++++++++}
>>>+cellidx(c,+'hello')
ans+=++1

>>>+cellidx(c,+'world')
ans+=++2|code-block|syntax|javascript|724370|entityMap|0|LINK|mutability|MUTABLE|url|http://www.obihiro.ac.jp/~suzukim/masuda/octave/html3/octave_36.html#SEC75^0|8|22|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

<p>Yes check this: <a href="http://www.obihiro.ac.jp/~suzukim/masuda/octave/html3/octave_36.html#SEC75">http://www.obihiro.ac.jp/~suzukim/masuda/octave/html3/octave_36.html#SEC75</a></p>

<pre><code>a = ["hello"; "world"];
c = cellstr (a)
     ⇒ c =
         {
           [1,1] = hello
           [2,1] = world
         }
&gt;&gt;&gt; cellidx(c, 'hello')
ans =  1

&gt;&gt;&gt; cellidx(c, 'world')
ans =  2
</code></pre>


blocks|key|507831|text|GNU+Octave在线性时间O(n)中搜索字符串单元格数组|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|507832|另一个答案是cellidx贬值了八度，它仍然可以运行，但他们说应该使用ismember，如下所示：|507833|%25linear+time+string+index+search.
a+=+["hello";+"unsorted";+"world";+"moobar"]
b+=+cellstr(a)
%25b+=
%25{
%25++[1,1]+=+hello
%25++[2,1]+=+unsorted
%25++[3,1]+=+world
%25++[4,1]+=+moobar
%25}
find(ismember(b,+'world'))+++%25returns+3|code-block|syntax|javascript|507834|“‘world”在索引槽3中。这是一个开销很大的linear+time+O(n)+operation，因为无论是否找到它，它都必须遍历所有元素。|507835|为了实现logarathmic+time+O(log+n)+solution，你的列表需要预先排序，然后你可以使用二进制搜索：|507836|如果你的单元格数组已经排序，你可以在最坏的情况下执行O(log-n)：|507837|function+i+=+binsearch(array,+val,+low,+high)
++%25binary+search+algorithm+for+numerics,+Usage:
++%25myarray+=+[+30,+40,+50.15+];++++++++%25already+sorted+list
++%25binsearch(myarray,+30,++++1,+3)+++++%25item+30+is+in+slot+1
++if+(+high+<+low+)
++++i+=+0;
++else
++++mid+=+floor((low+%2B+high)+/+2);
++++if+(+array(mid)+>+val+)
++++++i+=+binsearch(array,+val,+low,+mid-1);
++++elseif+(+array(mid)+<+val+)
++++++i+=+binsearch(array,+val,+mid%2B1,+high);
++++else
++++++i+=+mid;
++++endif
++endif
endfunction

function+i+=+binsearch_str(array,+val,+low,+high)
++%25+binary+search+for+strings,+usage:
++%25myarray2+=+[+"abc";+"def";+"ghi"];+++++++#already+sorted+list
++%25binsearch_str(myarray2,+"abc",+1,+3)+++++#item+abc+is+in+slot+1
++if+(+high+<+low+)
++++i+=+0;
++else
++++mid+=+floor((low+%2B+high)+/+2);
++++if+(+mystrcmp(array(mid,+[1:end]),+val)+==+1+)
++++++i+=+binsearch(array,+val,+low,+mid-1);
++++elseif+(+mystrcmp(array(mid,+[1:end]),+val)+==+-1+)
++++++i+=+binsearch_str(array,+val,+mid%2B1,+high);
++++else
++++++i+=+mid;
++++endif
++endif
endfunction
function+ret+=+mystrcmp(a,+b)
++++%25this+function+is+just+an+octave+string+compare,+it's+exactly+like+
++++%25strcmp(str1,str2)+in+C,+or+java.lang.String.compareTo(...)+in+Java.
++++%25returns+1+if+string+a+>+b
++++%25returns+0+if+string+a+==+b
++++%25return+-1+if+string+a+<+b
++++letters_gt+=+gt(a,+b);++++++%25list+of+boolean+a+>+b
++++letters_lt+=+lt(a,+b);++++++%25list+of+boolean+a+<+b
++++ret+=+0;
++++%25octave+makes+us+roll+our+own+string+compare+because+
++++%25strings+are+arrays+of+numerics
++++len+=+length(letters_gt);
++++for+i+=+1:len
++++++++if+letters_gt(i)+>+letters_lt(i)
++++++++++++ret+=+1;
++++++++++++return
++++++++elseif+letters_gt(i)+<+letters_lt(i)
++++++++++++ret+=+-1;
++++++++++++return
++++++++endif
++++end;
endfunction

%25Assuming+that+myarray+is+already+sorted,+(it+must+be+for+binary+
%25search+to+finish+in+logarithmic+time+`O(log-n))`+worst+case,+then+do

myarray+=+[+30,+40,+50.15+];++++++++%25already+sorted+list
binsearch(myarray,+30,++++1,+3)+++++%25item+30+is+in+slot+1
binsearch(myarray,+40,++++1,+3)+++++%25item+40+is+in+slot+2
binsearch(myarray,+50,++++1,+3)+++++%2550+does+not+exist+so+return+0
binsearch(myarray,+50.15,+1,+3)+++++%2550.15+is+in+slot+3

%25same+but+for+strings:
myarray2+=+[+"abc";+"def";+"ghi"];+++++++%25already+sorted+list
binsearch_str(myarray2,+"abc",+1,+3)+++++%25item+abc+is+in+slot+1
binsearch_str(myarray2,+"def",+1,+3)+++++%25item+def+is+in+slot+2
binsearch_str(myarray2,+"zzz",+1,+3)+++++%25zzz+does+not+exist+so+return+0
binsearch_str(myarray2,+"ghi",+1,+3)+++++%25item+ghi+is+in+slot+3|507838|若要对数组进行排序，请执行以下操作：|507839|排序的复杂性取决于您拥有的数据类型以及GNU+octave语言作者选择的排序算法，它介于O(n*log(n))和O(n*n)之间。|507840|myarray+=+[+9,+40,+-3,+3.14,+20+];++++++++%25not+sorted+list+
myarray+=+sort(myarray)+

myarray2+=+[+"the";+"cat";+"sat";+"on";+"the";+"mat"];+++++++%25not+sorted+list+
myarray2+=+sortrows(myarray2)|507841|不要在胶带上羞涩，这是维系整个部队的唯一力量。|507842|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Linear_time#Linear_time|1|https://stackoverflow.com/a/2307314/445131^0|F|4|0|6|7|Z|8|0|0|O|Q|0|0|4|Y|1|0|Q|8|0|0|0|18|B|1K|6|0|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1F|8|@$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|1K|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|1L|8|@]|D|@$9|1M|A|1N|1|1O]]|E|$]]|$1|O|3|P|5|6|7|1P|8|@]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|Q|3|R|5|6|7|1T|8|@$9|1U|A|1V|B|C]]|D|@]|E|$]]|$1|S|3|T|5|J|7|1W|8|@]|D|@]|E|$K|L]]|$1|U|3|V|5|6|7|1X|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1Y|8|@$9|1Z|A|20|B|C]|$9|21|A|22|B|C]]|D|@]|E|$]]|$1|Y|3|Z|5|J|7|23|8|@]|D|@]|E|$K|L]]|$1|10|3|11|5|6|7|24|8|@]|D|@]|E|$]]|$1|12|3|-4|5|6|7|25|8|@]|D|@]|E|$]]]|13|$14|$5|15|16|17|E|$18|19]]|1A|$5|15|16|17|E|$18|1B]]]]

<h2>GNU Octave search a cell array of strings in linear time <code>O(n)</code>:</h2>

<p>The other answer has <code>cellidx</code> which is depreciated by octave, it still runs but they say to use <code>ismember</code> instead, like this:</p>

<pre><code>%linear time string index search.
a = ["hello"; "unsorted"; "world"; "moobar"]
b = cellstr(a)
%b =
%{
%  [1,1] = hello
%  [2,1] = unsorted
%  [3,1] = world
%  [4,1] = moobar
%}
find(ismember(b, 'world'))   %returns 3
</code></pre>

<p>'world' is in index slot 3.  This is a expensive <a href="http://en.wikipedia.org/wiki/Linear_time#Linear_time" rel="noreferrer">linear time O(n) operation</a> because it has to iterate through all elements whether or not it is found.</p>

<p>To achieve a <a href="https://stackoverflow.com/a/2307314/445131">logarathmic time O(log n) solution</a>, then your list needs to come pre-sorted and then you can use binary search:</p>

<h2>If your cell array is already sorted, you can do <code>O(log-n)</code> worst case:</h2>

<pre><code>function i = binsearch(array, val, low, high)
  %binary search algorithm for numerics, Usage:
  %myarray = [ 30, 40, 50.15 ];        %already sorted list
  %binsearch(myarray, 30,    1, 3)     %item 30 is in slot 1
  if ( high &lt; low )
    i = 0;
  else
    mid = floor((low + high) / 2);
    if ( array(mid) &gt; val )
      i = binsearch(array, val, low, mid-1);
    elseif ( array(mid) &lt; val )
      i = binsearch(array, val, mid+1, high);
    else
      i = mid;
    endif
  endif
endfunction

function i = binsearch_str(array, val, low, high)
  % binary search for strings, usage:
  %myarray2 = [ "abc"; "def"; "ghi"];       #already sorted list
  %binsearch_str(myarray2, "abc", 1, 3)     #item abc is in slot 1
  if ( high &lt; low )
    i = 0;
  else
    mid = floor((low + high) / 2);
    if ( mystrcmp(array(mid, [1:end]), val) == 1 )
      i = binsearch(array, val, low, mid-1);
    elseif ( mystrcmp(array(mid, [1:end]), val) == -1 )
      i = binsearch_str(array, val, mid+1, high);
    else
      i = mid;
    endif
  endif
endfunction
function ret = mystrcmp(a, b)
    %this function is just an octave string compare, it's exactly like 
    %strcmp(str1,str2) in C, or java.lang.String.compareTo(...) in Java.
    %returns 1 if string a &gt; b
    %returns 0 if string a == b
    %return -1 if string a &lt; b
    letters_gt = gt(a, b);      %list of boolean a &gt; b
    letters_lt = lt(a, b);      %list of boolean a &lt; b
    ret = 0;
    %octave makes us roll our own string compare because 
    %strings are arrays of numerics
    len = length(letters_gt);
    for i = 1:len
        if letters_gt(i) &gt; letters_lt(i)
            ret = 1;
            return
        elseif letters_gt(i) &lt; letters_lt(i)
            ret = -1;
            return
        endif
    end;
endfunction

%Assuming that myarray is already sorted, (it must be for binary 
%search to finish in logarithmic time `O(log-n))` worst case, then do

myarray = [ 30, 40, 50.15 ];        %already sorted list
binsearch(myarray, 30,    1, 3)     %item 30 is in slot 1
binsearch(myarray, 40,    1, 3)     %item 40 is in slot 2
binsearch(myarray, 50,    1, 3)     %50 does not exist so return 0
binsearch(myarray, 50.15, 1, 3)     %50.15 is in slot 3

%same but for strings:
myarray2 = [ "abc"; "def"; "ghi"];       %already sorted list
binsearch_str(myarray2, "abc", 1, 3)     %item abc is in slot 1
binsearch_str(myarray2, "def", 1, 3)     %item def is in slot 2
binsearch_str(myarray2, "zzz", 1, 3)     %zzz does not exist so return 0
binsearch_str(myarray2, "ghi", 1, 3)     %item ghi is in slot 3
</code></pre>

<h2>To sort your array if it isn't already:</h2>

<p>Complexity of sorting depends on the kind of data you have and whatever sorting algorithm GNU octave language writers selected, it's somewhere between <code>O(n*log(n))</code> and <code>O(n*n)</code>.  </p>

<pre><code>myarray = [ 9, 40, -3, 3.14, 20 ];        %not sorted list 
myarray = sort(myarray) 

myarray2 = [ "the"; "cat"; "sat"; "on"; "the"; "mat"];       %not sorted list 
myarray2 = sortrows(myarray2)
</code></pre>

<p>Don't be shy on the duct tape with this dob, it's the only thing holding the unit together.</p>


<p>Is there built-in functionality for this?</p>


问如何在Octave中对单元格数组(字符串)中的元素进行排序和高效查找？
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问如何在Octave中对单元格数组(字符串)中的元素进行排序和高效查找？
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