我有一個矩陣，如：查找子矩陣
df<-data.frame(a=c(1,2,5,4,5,4), b=c(3,4,8,6,7,4))
，我想知道，如果下面的基質被包含在前面的一個地方：
df1<-data.frame(a=c(5,4), b=c(7,4))
我知道如何尋找一個元素：
which(df ==df1[1,1], arr.ind=T)
但不是完全矩陣。我需要獲取大矩陣中的子矩陣的座標。在這種情況下將是
(5,1;6,2)
有沒有辦法解決這個問題，而不必做循環？
我不知道，如果你的答案(3, 1; 4, 2)是正確的，但是這是我想出瞭解決方案：
mapply(function(x, y) which(x %in% y), df, df1)
a b
[1,] 3 2
[2,] 5 4
我真的不認爲有一種方法來避免環路要誠實：
# find all matches of the top left corner of df1
hits <- which(df==df1[1,1],arr.ind=TRUE)
# remove those matches that can't logically fit in the data
hits <- hits[hits[,"row"] <= nrow(df)-nrow(df1)+1,,drop=FALSE]
# check which of the matches is a hit...
# returning the top left corner of where the match is
hits[apply(
hits,
1,
function(x)
all(df[matrix(c(x,x+1:0,x+0:1,x+1),ncol=2,byrow=TRUE)] == unlist(df1))
)]
#[1] 5 1
不錯的主意，但不能用marix重複的子矩陣的第一個數字ans仍然有行來完成維度。 – GabyLP
Rcpp是這類問題的好工具。
我有點過分在這裏寫了一個非常複雜的函數，它可以找到較大數組中較小數組的所有匹配項的最小索引（對於矩陣是左上角）的座標，對於任何維度。如果你想在11維數組中找到所有9維數組的位置，這個函數可以爲你做。
這就是：
library('Rcpp');
cppFunction('
IntegerMatrix findarray(IntegerVector big, IntegerVector small, bool nacmp=true) {
// debugging macros
#define QUOTEID(...) #__VA_ARGS__
#define QUOTE(...) QUOTEID(__VA_ARGS__)
#define PRINT_VEC(vec,...) Rprintf(QUOTE(vec)"={"); if (vec.size() > 0) { Rprintf("%ld",vec[0]); for (size_t i = 1; i < vec.size(); ++i) Rprintf(",%ld",vec[i]); } Rprintf("}"__VA_ARGS__);
typedef std::vector<size_t> Dims;
// get big dimensions, treating a plain vector as a 1D array
Dims bigdims;
SEXP bigdimsSE = big.attr("dim");
if (Rf_isNull(bigdimsSE)) {
bigdims.push_back(big.size());
} else {
bigdims = as<Dims>(bigdimsSE);
}
//PRINT_VEC(bigdims,"\\n");
// now we can use this macro to easily return a result matrix with no matches
#define RES_NOMATCH IntegerMatrix(0,bigdims.size())
// get small dimensions, treating a plain vector as a 1D array
Dims smalldims;
SEXP smalldimsSE = small.attr("dim");
if (Rf_isNull(smalldimsSE)) {
smalldims.push_back(small.size());
} else {
smalldims = as<Dims>(smalldimsSE);
}
//PRINT_VEC(smalldims,"\\n");
// trivial case: if small has greater dimensionality than big, just return no matches
// note: we could theoretically support this case, at least when all extra small dimensions have only one index, but whatever
if (smalldims.size() > bigdims.size())
return RES_NOMATCH;
// derive a "bounds" Dims object, which will represent the maximum index plus one in big against which we must compare the first index in small for the corresponding dimension
// if small is greater than big in any dimension, then we can return no matches immediately
Dims bounds(smalldims.size());
for (size_t i = 0; i < smalldims.size(); ++i) {
if (smalldims[i] > bigdims[i])
return RES_NOMATCH;
bounds[i] = bigdims[i]-smalldims[i]+1;
}
// trivial case: if either big or small has any zero-length dimension, then just return no matches, because in that case the offending argument cannot have any actual data in it
// theoretically you can consider such degenerate arrays to match everywhere, sort of like the empty string matching at every position in any given string, but whatever
for (size_t i = 0; i < bigdims.size(); ++i) if (bigdims[i] == 0) return RES_NOMATCH;
for (size_t i = 0; i < smalldims.size(); ++i) if (smalldims[i] == 0) return RES_NOMATCH;
// prepare to build up the result data
// it would not make sense to build up the result data directly in a matrix, because we have to add one row at a time, which does not commute with the internal storage arrangement of matrices
// I then tried to use a data.frame, but the Rcpp DataFrame type is surprisingly light in functionality, seemingly without any provision for adding a row, and requires named columns, so best to avoid that
// instead, we\'ll just build up the data on a vector of vectors, going all-STL
typedef std::vector<std::vector<int> > ResBuilder;
ResBuilder resBuilder(bigdims.size());
// retrieve raw vector pointers for best performance
int* bigp = INTEGER(big);
int* smallp = INTEGER(small);
// now, iterate through each index of each (big) dimension from zero through the bound for that dimension (which is automatically the big dimension\'s length if small\'s dimensionality does not extend to that dimension), and see if small\'s first element matches
Dims bdis(bigdims.size()); // conveniently, initializes to all zeroes
size_t bvi = 0; // big vector index
while (true) { // big element loop, restricted to bounds
if (bigp[bvi] == smallp[0] && (nacmp || bigp[bvi] != NA_INTEGER)) {
//PRINT_VEC(bdis," ") Rprintf("found first element match at bvi=%ld big=small=%d\\n",bvi,bigp[bvi]);
size_t bvi2 = bvi; // don\'t screw up the original bvi; matches can overlap
// now we need to iterate through each index of each (small) dimension and test if all remaining elements match
Dims sdis(smalldims.size()); // conveniently, initializes to all zeroes
size_t svi = 0;
bool match = true; // assumption
while (true) { // small element loop
// note: once inside this inner loop, we don\'t have to worry about bounds anymore, because we already enforced that the outer loop will only iterate over indexes within bounds
// increment small and big indexes
++svi; // always increment svi by exactly one; the small array governs this matching loop
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("incremented svi=%ld\\n",svi);
size_t bm = 1;
size_t d;
for (d = 0; d < sdis.size(); ++d) {
++sdis[d];
++bvi2;
if (sdis[d] == smalldims[d]) {
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("reached small end=%ld of dimension d=%ld; bvi2=%ld bm=%ld\\n",smalldims[d],d,bvi2,bm);
sdis[d] = 0;
bvi2 += (bigdims[d]-smalldims[d])*bm-1;
bm *= bigdims[d];
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("after jumping to next index we have bvi2=%ld bm=%ld\\n",bvi2,bm);
} else {
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("valid dimension index increment at dimension d=%ld; bvi2=%ld bm=%ld\\n",d,bvi2,bm);
break;
}
}
// test if we reached the end of small; then break the inner while loop, and we have a match
if (d == sdis.size())
break;
// at this point, we have a new element to test; if unequal, we have no match
if (bigp[bvi2] != smallp[svi] || !nacmp && bigp[bvi] == NA_INTEGER) {
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("match overturned by big=%d != small=%d\\n",bigp[bvi2],smallp[svi]);
match = false;
break;
} else {
//PRINT_VEC(bdis," ") PRINT_VEC(sdis," ") Rprintf("match respected by big=small=%d\\n",bigp[bvi2]);
}
}
// if we have a match, add it to the result data
if (match) {
//PRINT_VEC(bdis," ") Rprintf("found complete match!\\n");
for (size_t bd = 0; bd < bigdims.size(); ++bd)
resBuilder[bd].push_back(bdis[bd]+1); // also add one to convert from C++ zero-based to R one-based indexes
//PRINT_VEC(bdis," ") Rprintf("resBuilder dims = {%ld,%ld}\\n",resBuilder[0].size(),resBuilder.size());
}
} else {
//PRINT_VEC(bdis," ") Rprintf("first element mismatch: big=%d != small=%d\\n",bigp[bvi],smallp[0]);
}
// increment big index
size_t bm = 1;
size_t d;
for (d = 0; d < bdis.size(); ++d) {
++bdis[d];
++bvi;
size_t bound = bounds.size() > d ? bounds[d] : bigdims[d];
if (bdis[d] >= bound) {
//PRINT_VEC(bdis," ") Rprintf("big index hit bound=%ld of dimension d=%ld; bvi=%ld bm=%ld\\n",bound,d,bvi,bm);
bdis[d] = 0;
bvi += (bigdims[d]-bound)*bm-1;
bm *= bigdims[d];
//PRINT_VEC(bdis," ") Rprintf("after advancing big index we have bvi=%ld bm=%ld\\n",bvi,bm);
} else {
//PRINT_VEC(bdis," ") Rprintf("valid dimension index increment at dimension d=%ld; bvi=%ld bm=%ld\\n",d,bvi,bm);
break;
}
}
// test if we reached the end of big; then break the outer while loop, and we\'re done
if (d == bdis.size() || bvi >= big.size())
break;
}
// copy to a matrix
IntegerMatrix res(resBuilder[0].size(),resBuilder.size());
int* resp = INTEGER(res);
for (size_t c = 0; c < res.ncol(); ++c)
std::copy(resBuilder[c].begin(),resBuilder[c].end(),resp+c*res.nrow());
// return the matrix
return res;
}
');
這是我做了一些非常隨意的測試，僅達立方體的立方體（每個測試打印的big陣列，那麼small數組，然後將結果，最後邏輯向量測試如果small在big從每個連續的比賽延長大小的切片真的等同於small）：
## testing
slice <- function(arr,is,ls,...) { length(ls) <- length(is); ls[is.na(ls)] <- 1; do.call(`[`,c(list(arr),Map(function(i,l) seq(i,len=l),is,ls),...)); };
printAndTest <- function(big,small) { print(big); print(small); findarray(big,small); };
printAndTestAndSliceIdentical <- function(big,small) { big <- structure(as.integer(big),dim=dim(big)); small <- structure(as.integer(small),dim=dim(small)); res <- printAndTest(big,small); print(res); if (nrow(res) > 0) sapply(1:nrow(res),function(r) identical(structure(slice(big,res[r,],if (is.null(dim(small))) length(small) else dim(small),drop=F),dim=dim(small)),small)) else logical(); };
## one-element match
printAndTestAndSliceIdentical(1,1);
## [1] 1
## [1] 1
## [,1]
## [1,] 1
## [1] TRUE
## vector in vector
printAndTestAndSliceIdentical(1:3,2:3);
## [1] 1 2 3
## [1] 2 3
## [,1]
## [1,] 2
## [1] TRUE
printAndTestAndSliceIdentical(1:3,1:3);
## [1] 1 2 3
## [1] 1 2 3
## [,1]
## [1,] 1
## [1] TRUE
printAndTestAndSliceIdentical(1:3,1:4);
## [1] 1 2 3
## [1] 1 2 3 4
## [,1]
## logical(0)
## vector in matrix
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),1:2);
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [1] 1 2
## [,1] [,2]
## [1,] 1 1
## [2,] 1 4
## [1] TRUE TRUE
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),12);
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [1] 12
## [,1] [,2]
## [1,] 4 3
## [2,] 4 6
## [1] TRUE TRUE
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),5:8);
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [1] 5 6 7 8
## [,1] [,2]
## [1,] 1 2
## [2,] 1 5
## [1] TRUE TRUE
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),5:9);
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [1] 5 6 7 8 9
## [,1] [,2]
## logical(0)
## matrix in matrix
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),matrix(1:4,2));
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [,1] [,2]
## [1,] 1 3
## [2,] 2 4
## [,1] [,2]
## logical(0)
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),matrix(c(2,3,6,7),2));
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [,1] [,2]
## [1,] 2 6
## [2,] 3 7
## [,1] [,2]
## [1,] 2 1
## [2,] 2 4
## [1] TRUE TRUE
printAndTestAndSliceIdentical(matrix(rep(1:12,2),4),matrix(c(7,8,11,12),2));
## [,1] [,2] [,3] [,4] [,5] [,6]
## [1,] 1 5 9 1 5 9
## [2,] 2 6 10 2 6 10
## [3,] 3 7 11 3 7 11
## [4,] 4 8 12 4 8 12
## [,1] [,2]
## [1,] 7 11
## [2,] 8 12
## [,1] [,2]
## [1,] 3 2
## [2,] 3 5
## [1] TRUE TRUE
## vector in cube
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),1);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 1
## [,1] [,2] [,3]
## [1,] 1 1 1
## [2,] 1 1 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),8);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 8
## [,1] [,2] [,3]
## [1,] 4 2 1
## [2,] 4 2 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),9);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 9
## [,1] [,2] [,3]
## [1,] 1 3 1
## [2,] 1 3 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),12);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 12
## [,1] [,2] [,3]
## [1,] 4 3 1
## [2,] 4 3 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),1:4);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 1 2 3 4
## [,1] [,2] [,3]
## [1,] 1 1 1
## [2,] 1 1 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),1:5);
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [1] 1 2 3 4 5
## [,1] [,2] [,3]
## logical(0)
## matrix in cube
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),matrix(c(7,8,11,12),2));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [,1] [,2]
## [1,] 7 11
## [2,] 8 12
## [,1] [,2] [,3]
## [1,] 3 2 1
## [2,] 3 2 2
## [1] TRUE TRUE
printAndTestAndSliceIdentical(array(1:12,c(4,3,2)),matrix(c(7,8,11,11),2));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## [,1] [,2]
## [1,] 7 11
## [2,] 8 11
## [,1] [,2] [,3]
## logical(0)
## cube in cube
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(c(1,13,25),c(1,1,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1]
## [1,] 1
##
## , , 2
##
## [,1]
## [1,] 13
##
## , , 3
##
## [,1]
## [1,] 25
##
## [,1] [,2] [,3]
## [1,] 1 1 1
## [1] TRUE
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(c(6,18,30),c(1,1,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1]
## [1,] 6
##
## , , 2
##
## [,1]
## [1,] 18
##
## , , 3
##
## [,1]
## [1,] 30
##
## [,1] [,2] [,3]
## [1,] 2 2 1
## [1] TRUE
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(c(18,30),c(1,1,2)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1]
## [1,] 18
##
## , , 2
##
## [,1]
## [1,] 30
##
## [,1] [,2] [,3]
## [1,] 2 2 2
## [1] TRUE
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(1:36,c(4,3,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## [,1] [,2] [,3]
## [1,] 1 1 1
## [1] TRUE
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(c(7,8,11,12,19,20,23,24,31,32,35,36),c(2,2,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1] [,2]
## [1,] 7 11
## [2,] 8 12
##
## , , 2
##
## [,1] [,2]
## [1,] 19 23
## [2,] 20 24
##
## , , 3
##
## [,1] [,2]
## [1,] 31 35
## [2,] 32 36
##
## [,1] [,2] [,3]
## [1,] 3 2 1
## [1] TRUE
printAndTestAndSliceIdentical(array(1:36,c(4,3,3)),array(c(7,8,11,12,19,20,23,24,31,32,35,37),c(2,2,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1] [,2]
## [1,] 7 11
## [2,] 8 12
##
## , , 2
##
## [,1] [,2]
## [1,] 19 23
## [2,] 20 24
##
## , , 3
##
## [,1] [,2]
## [1,] 31 35
## [2,] 32 37
##
## [,1] [,2] [,3]
## logical(0)
printAndTestAndSliceIdentical(array(1:36,c(4,3,6)),array(c(7,8,11,12,19,20,23,24,31,32,35,36),c(2,2,3)));
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 4
##
## [,1] [,2] [,3]
## [1,] 1 5 9
## [2,] 2 6 10
## [3,] 3 7 11
## [4,] 4 8 12
##
## , , 5
##
## [,1] [,2] [,3]
## [1,] 13 17 21
## [2,] 14 18 22
## [3,] 15 19 23
## [4,] 16 20 24
##
## , , 6
##
## [,1] [,2] [,3]
## [1,] 25 29 33
## [2,] 26 30 34
## [3,] 27 31 35
## [4,] 28 32 36
##
## , , 1
##
## [,1] [,2]
## [1,] 7 11
## [2,] 8 12
##
## , , 2
##
## [,1] [,2]
## [1,] 19 23
## [2,] 20 24
##
## , , 3
##
## [,1] [,2]
## [1,] 31 35
## [2,] 32 36
##
## [,1] [,2] [,3]
## [1,] 3 2 1
## [2,] 3 2 4
## [1] TRUE TRUE
這裏對您的數據演示：
df <- data.frame(a=c(1,2,5,4,5,4),b=c(3,4,8,6,7,4));
df1 <- data.frame(a=c(5,4),b=c(7,4));
findarray(as.matrix(df),as.matrix(df1));
## [,1] [,2]
## [1,] 5 1
我的函數只返回最低指數的座標，因爲你可以得到的最高指數的座標通過簡單地添加的small大小，如下所示：
t(t(findarray(as.matrix(df),as.matrix(df1)))+dim(df1))-1;
## [,1] [,2]
## [1,] 6 2
注意，換位是必要的因爲R對較大矩陣（即，跨行，然後跨列）。對於您的特定數據，這顯然不是必需的，因爲只有一個匹配，並且此外，兩個維度的長度都相同，所以它無關緊要，但在一般情況下很重要。
好了，就這樣我可以說我做了，這裏是在11D陣列匹配9D的一個簡單的測試：
set.seed(12);
big <- array(sample(1:4,factorial(11),replace=T),11:1);
small <- array(sample(1:4,12,replace=T),c(2,3,2,rep(1,9-3)));
res <- findarray(big,small);
res;
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## [1,] 6 6 5 3 1 3 4 3 2 1 1
## [2,] 7 7 6 3 5 6 5 4 3 2 1
sapply(1:nrow(res),function(r) identical(structure(slice(big,res[r,],dim(small),drop=F),dim=dim(small)),small));
## [1] TRUE TRUE
思想，以測試這另一個好辦法：我們可以從大陣中取出片，看看findarray()是否可以找到它們。
set.seed(96);
d <- 11;
big <- array(sample(1:4,factorial(d),replace=T),d:1);
for (i in 1:5) {
is <- sapply(d:1,sample,1);
ls <- mapply(function(i,dl) sample(dl-i+1,1),is,d:1);
small <- slice(big,is,ls,drop=F);
res <- findarray(big,small);
print(rbind(is,ls,res));
};
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## is 7 6 1 4 7 2 2 3 3 1 1
## ls 3 1 2 1 1 1 1 2 1 1 1
## 5 3 6 8 4 4 4 2 1 1 1
## 7 6 1 4 7 2 2 3 3 1 1
## 8 10 7 5 1 2 2 3 1 2 1
## 9 6 3 4 4 1 4 3 3 2 1
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## is 10 10 2 4 5 6 3 1 3 2 1
## ls 2 1 3 4 1 1 3 1 1 1 1
## 10 10 2 4 5 6 3 1 3 2 1
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## is 8 5 5 8 2 1 5 4 1 1 1
## ls 2 1 1 1 2 3 1 1 1 1 1
## 8 5 5 8 2 1 5 4 1 1 1
## 1 4 3 1 5 1 2 1 3 1 1
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## is 7 10 7 7 6 3 5 4 3 2 1
## ls 2 1 1 2 2 2 1 1 1 1 1
## 7 10 7 7 6 3 5 4 3 2 1
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
## is 3 8 5 1 6 3 1 3 3 2 1
## ls 9 1 2 7 2 3 4 1 1 1 1
## 3 8 5 1 6 3 1 3 3 2 1
你應該停止稱他們爲矩陣如果你是問有關R. –
問題。您確定要DF1例子是代表你在做什麼。例如。 - 它不應該是'df1 <-data.frame（a = c（5,4），b = c（8,6））'？ – thelatemail