我有一个矩阵（5x10000），第五行包含1到50之间的值，对应于实验的不同事件。 我的目标是找到矩阵的列，这些列对于不同的事件是相同的。 换句话说，我希望不同事件的所有可能组合的列结果（{1,2，..，50}的子集）（例如：{1,3,7}和{7,1,3}是当然是相同的组合）。 这听起来像是集合的交集问题，每个集合都包含给定事件的所有可能结果。 我也希望计算时间合理。
矩阵示例（5x20）：
A =
20 4 4 74 20 20 3 1 1 4 3 3 3 7 4 1 20 3 3 74 36 1 1 11 36 36 3 3 3 1 3 3 3 9 4 3 36 4 3 11 77 1 1 15 77 77 1 3 3 1 1 1 1 10 3 2 77 4 1 15 9 4 4 40 9 9 2 4 4 4 2 2 2 40 1 4 9 3 2 40 3 4 2 6 7 3 4 5 2 7 4 2 7 6 7 2 5 5 1 3在这种情况下，我们有7个不同的事件，从1到7：第5行
例如：
事件3,5和7的结果的交集是向量：[20 36 77 9]'
事件1,2,4和7的结果的交集是向量：[3 3 1 2]'
事件3和6的结果的交集是向量：[20 36 77 9]'和[74 11 15 40]'
所以我想要的是1到50之间指定数量的不同事件的公共列。例如，如何获得20个不同事件的共同列？ 当我想在集合{1,2，...，50}中找到20个事件的所有可能组合的结果时，问题变得更加复杂。
对于给定数量的不同事件，我想要所有可能组合的公共列，但是我给出了数字20作为基于一个解决方案的示例。
我会重新提出我的问题，以便更清楚：
以下矩阵是A的子矩阵，每个矩阵对应一个给定的事件：
A1 = [3; 3; 1; 2; 1]
A1对应于事件1的结果
A2 = [4 1 3 1; 1 3 3 3; 1 3 1 2; 4 4 2 4; 2 2 2 2]
A2对应于事件2的结果
A3 = [20 20 74; 36 36 11; 77 77 15; 9 9 40; 3 3 3]
A3对应于事件3的结果
A4 = [4 3 3; 1 3 3; 1 1 1; 4 2 2; 4 4 4]
A4对应于事件4的结果
A5 = [1 20 3; 3 36 4; 3 77 4; 4 9 3; 5 5 5]
A5对应于事件5的结果
A6 = [74 7; 11 9; 15 10; 40 40; 6] 6
A6对应于事件6的结果
A7 = [20 4 3 4 7; 36 1 3 4; 77 1 1 3; 9 4 2 1; 7 7 7 7]
A7对应于事件的结果
我的目标是找到矩阵Ai（1：4，:) i = 1,2，... 7的列的交点
换一种说法：
交叉（Ai，Aj）（1：4，:)对于i和j不同
交叉（Ai，Aj，Ak）（1：4，:)对于i，j和k不同
对于i，j，k和l不同，交集（Ai，Aj，Ak，Al）（1：4，:)
交叉（Ai，Aj，Ak，Al，Am）（1：4，:)对于i，j，k，l和m不同
i，j，k，l，m和n不同的交叉点（Ai，Aj，Ak，Al，Am，An）（1：4，:)
i，j，k，l，m，n和o的交叉点（Ai，Aj，Ak，Al，Am，An，Ao）（1：4，:)
i，j，k，l，m，n，o和p不同的交叉点（Ai，Aj，Ak，Al，Am，An，Ao，Ap）（1：4，:)
当我说“交叉（Ai，Aj）（1：4，:)为i和j不同，”我想要矩阵Ai（1：4，:)和Aj（1：4，:)共有的列
每个交集的结果可以是许多列向量，不一定是一个，这取决于矩阵A的列。
我希望每个结果都包含矩阵Ai（1：4，:)的向量列，后跟相应的事件值，例如：if [3 3 1 2]'是A1，A2，A4的交集和A7，我想得到矢量[3 3 1 2 1 2 4 7]'
例如：十字路口（A1，A2，A3，A4）（1：4，:)：我的目标是避免以下循环：
[n1 m1] = size(A1); [n2 m2] = size(A2); [n3 m3] = size(A3); [n4 m4] = size(A4); k=1; for i1=1:m1 for i2=1:m2 for i3=1:m3 for i4=1:m4 if A1(1:4,i1)==A2(1:4,i2) && A2(1:4,i2)==A3(1:4,i3) && A3(1:4,i3)==A4(1:4,i4) intersection1234(:,k) = [A1(1:4,i1);A1(5,i1);A2(5,i2);A3(5,i3);A4(5,i4)]; k=k+1; end end end end endI have a matrix (5x10000) with the fifth line contains values between 1 and 50 corresponding to different events of an experiment. My goal is to find the columns of the matrix which are the same for different events. In other words, I want the columns results for all possible combinations of different events (subsets of {1,2, .., 50}) (For example: {1,3,7} and {7,1,3} are of course the same combination). It sounds like a problem of intersection of sets that each contains all possible outcomes for a given event. I hope also that the computation time is reasonable.
example with a matrix (5x20):
A =
20 4 4 74 20 20 3 1 1 4 3 3 3 7 4 1 20 3 3 74 36 1 1 11 36 36 3 3 3 1 3 3 3 9 4 3 36 4 3 11 77 1 1 15 77 77 1 3 3 1 1 1 1 10 3 2 77 4 1 15 9 4 4 40 9 9 2 4 4 4 2 2 2 40 1 4 9 3 2 40 3 4 2 6 7 3 4 5 2 7 4 2 7 6 7 2 5 5 1 3in this case we have seven different events from 1 to 7: line 5
for example:
the intersection of the results of events 3, 5 and 7 is the vector: [20 36 77 9]'
the intersection of the results of events 1, 2, 4 and 7 is the vector: [3 3 1 2]'
the intersection of the results of events 3 and 6 are the vectors: [20 36 77 9]' and [74 11 15 40]'
So what I want is the common columns for a specified number of different events between 1 and 50. For example, how to get the columns common to 20 different events? The problem becomes more complicated for me when I think to find this result for all possible combinations of 20 events in the set {1,2,..., 50}.
I want the common columns for all possible combination for a given number of different events, but I gave the number 20 just as an example on which to base one solution.
I'll rephrase my question that to make it clearer:
the following matrices are sub-matrix of A, each corresponding to a given event:
A1= [3;3;1;2;1]
A1 corresponds to results of the event 1
A2= [4 1 3 1;1 3 3 3;1 3 1 2;4 4 2 4;2 2 2 2]
A2 corresponds to results of the event 2
A3= [20 20 74;36 36 11;77 77 15;9 9 40;3 3 3]
A3 corresponds to results of the event 3
A4= [4 3 3;1 3 3;1 1 1;4 2 2;4 4 4]
A4 corresponds to results of the event 4
A5= [1 20 3;3 36 4;3 77 4;4 9 3;5 5 5]
A5 corresponds to results of the event 5
A6= [74 7;11 9;15 10;40 40;6 ]6
A6 corresponds to results of the event 6
A7= [20 4 3 4 7;36 1 3 4;77 1 1 3;9 4 2 1;7 7 7 7]
A7 corresponds to results of the event
my goal is to find intersection along the columns of the matrix Ai (1:4,:) i = 1,2, ... 7
in other words:
intersection(Ai,Aj)(1:4,:) for i and j different
intersection(Ai,Aj,Ak)(1:4,:) for i,j and k different
intersection(Ai,Aj,Ak,Al)(1:4,:) for i,j,k and l different
intersection(Ai,Aj,Ak,Al,Am)(1:4,:) for i,j,k,l and m different
intersection(Ai,Aj,Ak,Al,Am,An)(1:4,:) for i,j,k,l,m and n different
intersection(Ai,Aj,Ak,Al,Am,An,Ao)(1:4,:) for i,j,k,l,m,n and o different
intersection(Ai,Aj,Ak,Al,Am,An,Ao,Ap)(1:4,:) for i,j,k,l,m,n,o and p different
when I say "intersection (Ai, Aj)(1:4,:) for i and j different," I want the columns common to the matrix Ai(1:4,:) and Aj(1:4,:)
the result for each intersection can be many column vectors, not necessarily one, depending on the columns of the matrix A.
I hope that each result contains the vector column of the matrix Ai(1:4,:) followed by the corresponding values of events, such as: if [3 3 1 2]' is the intersection of A1, A2, A4 and A7, I want to get as a result the vector [3 3 1 2 1 2 4 7]'
for example: intersection(A1,A2,A3,A4)(1:4,:): my goal is to avoid the following loop:
[n1 m1] = size(A1); [n2 m2] = size(A2); [n3 m3] = size(A3); [n4 m4] = size(A4); k=1; for i1=1:m1 for i2=1:m2 for i3=1:m3 for i4=1:m4 if A1(1:4,i1)==A2(1:4,i2) && A2(1:4,i2)==A3(1:4,i3) && A3(1:4,i3)==A4(1:4,i4) intersection1234(:,k) = [A1(1:4,i1);A1(5,i1);A2(5,i2);A3(5,i3);A4(5,i4)]; k=k+1; end end end end end最满意答案
如果我理解正确，您希望查找事件不同的列。 以John Colby的答案为基础：
n = 1e3 tic % Simulate data % (Here we've split off the 5th row into a separate variable) data = randi(5, [4 n]); exptEvents = randi(50, 1, n); % Find repeats [b,i,j] = unique(data', 'rows'); % Organize the indices of the repeated columns into a cell array reps = arrayfun(@(x) find(j==x), 1:length(i), 'UniformOutput', false); % Find events corresponding to these repeats reps_Events = cellfun(@(x) exptEvents(x), reps, 'UniformOutput', false); U = cellfun(@unique, reps_Events, 'UniformOutput', false); repeat_counts = cellfun(@length, U); k=20; rep_data = b(repeat_counts>=k,:); toc上面的代码中的U在每个单元格中具有唯一事件的组（或“组合”）。 每个单元格也对应于唯一的数据列。 如果您还需要其他产品，请举例说明。 rep_data包含在k个或更多事件中重复的结果。
If I understand correctly, you want to find columns whose events are different. Building on John Colby's answer:
n = 1e3 tic % Simulate data % (Here we've split off the 5th row into a separate variable) data = randi(5, [4 n]); exptEvents = randi(50, 1, n); % Find repeats [b,i,j] = unique(data', 'rows'); % Organize the indices of the repeated columns into a cell array reps = arrayfun(@(x) find(j==x), 1:length(i), 'UniformOutput', false); % Find events corresponding to these repeats reps_Events = cellfun(@(x) exptEvents(x), reps, 'UniformOutput', false); U = cellfun(@unique, reps_Events, 'UniformOutput', false); repeat_counts = cellfun(@length, U); k=20; rep_data = b(repeat_counts>=k,:); tocU in the code above has in every cell a group (or "combination") of unique events. Each cell also corresponds to a unique data column. If you need something else, please give an example. rep_data contains results that repeat in k or more events.
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