在八度/ MATLAB創建基於詞的存在,從一個矩陣的新矩陣,說我有:倍頻/ MATLAB:在另一個
all =
{
[1,1] = one
[1,2] = two
[1,3] = three
[1,4] = four
[1,5] = five
[1,6] = six
[1,7] = seven
[1,8] = eight
[1,9] = nine
[1,10] = ten
}
some =
{
[1,1] = someword
[1,2] = someword
[1,3] = one
[1,4] = someword
[1,5] = nine
}
我怎麼會做出新的矩陣,使得
new =
{
[1,1] = 1
[1,2] = 0
[1,3] = 0
[1,4] = 0
[1,5] = 0
[1,6] = 0
[1,7] = 0
[1,8] = 0
[1,9] = 1
[1,10] = 0
}
也就是說,new
矩陣的大小與all
矩陣的大小相同,但其值爲1
或0
,具體取決於中的單詞是否存在於all
?
我用你的第一個建議,它工作的很棒!謝啦!只是一個後續問題。如果我有另一個名爲'some2'的矩陣,並且我將相同的for-loop應用於它以得到1和0,我將如何將它添加到'new'矩陣,以便現在,'新'矩陣將有兩列 - 第一個包含1s和0s來自'some',第二個包含1s和0s來自'some2'? – Obay
您可以生成一個new2輸出,並使用'new_combined = new | new2;'從兩者中獲取非零值。 –
我得到'error:binary operator'|' 'cell'operations'not'for'cell' – Obay