(A, 0, 0, half_n, half_n) A12 = submatrix(A, 0, half_n, half_n, half_n) A21 = submatrix(A, half_n..., 0, half_n, half_n) A22 = submatrix(A, half_n, half_n, half_n, half_n) B11 = submatrix(B, 0, 0..., half_n, half_n) B12 = submatrix(B, 0, half_n, half_n, half_n) B21 = submatrix(B, half_n, 0, half_n...(C, 0, 0, C11) set_submatrix(C, 0, half_n, C12) set_submatrix(C, half_n, 0, C21) set_submatrix...(C, half_n, half_n, C22) return C请注意,此伪代码中使用的submatrix和set_submatrix函数是用于获取和设置矩阵的子矩阵的辅助函数。
eye(6) b = [1 0;0 1] disp(‘a矩阵中b的模的个数是:’); count = juZhenDeMo(a,b) end 求向量的模: function count = sta_submatrix1...end end clc; clear; a = [0 0 0 1 0 0 1 0 0 1 0 0 1 0 0] b = [0 0 ] disp(‘b在a中的模的个数是:’) count = sta_submatrix1
1, 2, 3], [4, 5, 6], [7, 8, 9]])# 创建一个子矩阵的索引indices = np.ix_([0, 1], [1, 2])# 使用ix_()函数将子矩阵转换为一个数组submatrix...= matrix[indices]# 打印结果print(submatrix)输出:[[2 3] [5 6]]使用这些方法可以大大提高代码的效率,并使其更易读和维护。
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/largest-submatrix-with-rearrangements 著作权归领扣网络所有
paradigm,SDP); 棋盘格刺激范式(checkerboard paradigm,CBP); 基于区域的刺激范式(regionbased paradigm,RBP); 基于子矩阵的刺激范式(submatrix...5.基于子矩阵的刺激范式(submatrix basedparadigm,SBP) ---- 具有四个3×3子矩阵的基于子矩阵的范例(SBP)。两条虚线将整个6×6键盘矩阵划分为四个3×3子矩阵。...columns 基于P300和SSVEP混合范式脑—机接口研究 A region-based P300 speller for brain-computer interface A dynamic submatrix-based
<= matrix.length, matrix[0].length <= 200 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/max-submatrix-lcci
Generalized Matrix Local Low Rank Representation by Random Projection and Submatrix Propagation 9....Generalized Matrix Local Low Rank Representation by Random Projection and Submatrix Propagation Pengtao
练习题如下: POJ 3250: Bad Hair Day POJ 2082: Terrible Sets POJ 3494: Largest Submatrix of All 1’s POJ 3250...POJ 3494: Largest Submatrix of All 1’s 都是些经典题,leetcode上也刷到过,利用上一题的做法,先统计当前行每一列对应的最大高度是多少,接着每一行都有对应的高度序列
x-direction * pulMap* - mappings of greylevels from histograms * uiXSize - uiXSize of image submatrix...* uiYSize - uiYSize of image submatrix * pLUT - lookup table containing mapping greyvalues...to bins * This function calculates the new greylevel assignments of pixels within a submatrix * of...mapTileRows(2), mapTileCols(2)}; % Calculate the new greylevel assignments of pixels % within a submatrix
row access A[[1]] as list: A[1] as 1xn matrix: row(A, 1) column access col(A, 1) submatrix...submatrix access How to access a submatrix.
avg$RNA b = as.data.frame(a) str(b) gene_boxplot <- function(gene){ average_expression <- colMeans(submatrix...} # Cytotoxic基因名称列表 Cytotoxic_gene <- c("GZMA", "GZMB", "GZMK" ,"GNLY", "IFNG" ,"PRF1" ,"NKG7") submatrix...gene_list Exhausted_gene <- c("LAG3" , "TIGIT" , "PDCD1" , "HAVCR2", "CTLA4" , "LAYN" , "ENTPD1") submatrix...assay = "RNA", group.by = "patientoutcone") a = avg$RNA b = as.data.frame(a) str(b) # Cytotoxic基因名称列表 submatrix...<- b[Cytotoxic_gene, ] gene_boxplot('Cytotoxic signature') # Exhausted基因名称列表 submatrix <- b[Exhausted_gene
注意:矩阵\(A\)是一个长方形矩阵,不一定是方阵,另外\(\Sigma\)和矩阵\(A\)的维度相同,并且其包含一个对角子矩阵(diagonal submatrix)。 2.
---->[mat.hpp#Mat::operator()]---- /** @overload @param roi Extracted submatrix specified as a rectangle
, 'MAT_AUTO_STEP', 'MAT_CONTINUOUS_FLAG', 'MAT_DEPTH_MASK', 'MAT_MAGIC_MASK', 'MAT_MAGIC_VAL', 'MAT_SUBMATRIX_FLAG..., 'Mat_AUTO_STEP', 'Mat_CONTINUOUS_FLAG', 'Mat_DEPTH_MASK', 'Mat_MAGIC_MASK', 'Mat_MAGIC_VAL', 'Mat_SUBMATRIX_FLAG...UMAT_DATA_TEMP_UMAT', 'UMAT_DATA_USER_ALLOCATED', 'UMAT_DEPTH_MASK', 'UMAT_MAGIC_MASK', 'UMAT_MAGIC_VAL', 'UMAT_SUBMATRIX_FLAG...UMat_AUTO_STEP', 'UMat_CONTINUOUS_FLAG', 'UMat_DEPTH_MASK', 'UMat_MAGIC_MASK', 'UMat_MAGIC_VAL', 'UMat_SUBMATRIX_FLAG
5.基于子矩阵的刺激范式(submatrix basedparadigm,SBP) ---- 具有四个3×3子矩阵的基于子矩阵的范例(SBP)。两条虚线将整个6×6键盘矩阵划分为四个3×3子矩阵。
|[BST, DFS, Tree]|| |359|Subarray Sum.java|Easy|Java|[Array, Hash Table, PreSum, Subarray]|| |360|Submatrix
pid=2845 横竖分别求一下不连续的最大子段和; 状态方程: Sum[i]=max(sum[j])+a[i];其中,0<=j<i-1; Largest Submatrix http://acm.hdu.edu.cn
]=max(f[i-1],f[i-2]+f[i]); } printf("%d\n",f[n]); } return 0; } 19.Largest Submatrix
}}; // 计算答案 int result = matrixSubmatrixSum(n, m, matrix); // 输出结果 printf("Sum of submatrix
with selected rectangle Mat submat = orig_img(Rect(p1, p2)); // Here some processing for the submatrix
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