Question

您好我正在尝试为imagej制作一个宏。现在我输入了一个大约300个元素的向量Array1y。如果我想在子向量中细分向量，我该怎么办？例如，如果我想要100个子向量，那么步骤为30，我就想到了类似的东西

w=30;
step= Array1y.length/w
for (i=0; i<Array1y.length; i+=(w+1)) 
{
for (j=w; j<Array1y.length ; j+=w)
a(i)= Array.slice(Array1y,i,w)
Array.print a(i)
}
}

怎么了？它可以工作，如果我没有300元素但更少，让我们说268，我也想要30步（所以我有最后一个不是由30个元素制作的子向量？）

Answer 1

我不确定您尝试做什么，或者实际上是java，但是我认为您要求的是：

public static void main(String[] args) {
    int[] vector = new int[268];
    IntStream.range(0, vector.length).forEach(i -> vector[i] = i); // to fill the vector
    int subvectorSize = 30;
    int[][] subvectors = new int[vector.length/subvectorSize+1][subvectorSize];  // array of arrays
    for(int i=0; i< vector.length; i++){  // iterate over vector
        int x = i/subvectorSize; // get quotient (xth subvector)
        int y = i%subvectorSize; // get reminder (index y of subvector)
        subvectors[x][y] = vector[i];
    }
    System.out.println(Arrays.deepToString(subvectors));
}

输出：

[[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22] ，23,24,25,26,27,28,29]，[30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46 ，47,48,49,50,51,52,53,54,55,56,57,58,59]，[60,61,62,63,64,65,66,67,68,69,70 ，71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89]，[90,91,92,93,94 ，95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119 ]，[120,121,122,123,124,125,126,127,128,129,130,131,132,133,134,135,136,137,138,139,140,141,142,143] ，144,145,146,147,148,149]，[150,151,152,153,154,155,156,157,158,159,160,161,162,163,164,165,166,167 ，168,169,170,171,172,173,174,175,176,177,178,179]，[180,181,182,183,184,185,186,187,188,189,190,191] ，192,193,194,195,196,197,198,199,200,201,202,203,204,205,206,207,208,209]，[210,211,212,213,214,215] ，216,217,2 18,219,220,221,222,223,224,225,226,227,228,229,230,231,232,233,234,235,236,237,238,239]，[240,241， 242,243,244,245,246,247,248,249,250,251,252,253,254,255,256,257,258,259,260,261,262,263,264,265,266， 267,0,0]]

如果循环，宏中的错误

1 个答案: