我有一个矩阵，如：
df<-data.frame(a=c(1,2,5,4,5,4), b=c(3,4,8,6,7,4))
我想知道前一个矩阵是否包含以下矩阵：
df1<-data.frame(a=c(5,4), b=c(7,4))
我知道如何寻找元素：
which( df ==df1[1,1], arr.ind=T )
但不是完全矩阵。我需要在大矩阵中获得子矩阵的坐标。在这种情况下将是
(5,1;6,2)
有没有办法在不进行循环的情况下解决这个问题？
答案 0 :(得分：5)
Rcpp是解决此类问题的好工具。
我有点过分了，写了一个非常复杂的函数，可以找到较大数组中较小数组的所有匹配的最低索引（对于矩阵的左上角）的坐标，对于任何维。如果你想在11维数组中找到9维数组的所有位置，这个函数可以为你做。
这是：
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;
}
');
这里有一些相当随意的测试，只有多维数据集在立方体中（每个测试打印big数组，然后是small数组，然后是结果，最后如果从small中的每个连续匹配延伸的big大小的切片与small完全相同，则进行逻辑矢量测试：
## 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
以下是您的数据演示：
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
我的函数只返回最低索引坐标，因为您可以通过简单地添加small的大小来导出最高索引坐标，如下所示：
t(t(findarray(as.matrix(df),as.matrix(df1)))+dim(df1))-1;
## [,1] [,2]
## [1,] 6 2
请注意，换位是必要的，因为R将短矢量与较大的矩阵（即跨行，然后跨列）循环。这显然不是您的特定数据所必需的，因为只有一个匹配，而且df1的两个维度具有相同的长度，所以无论如何它都不重要，但它在一般情况。
好吧，我可以说我做了，这是在11D阵列中匹配9D的简单测试：
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
考虑另一个测试方法的好方法：我们可以从大数组中获取切片，看看findarray()是否可以找到它们。
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
答案 1 :(得分：3)
我真的不认为有办法避免循环诚实：
# 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
答案 2 :(得分：0)
我不知道您的回答(3, 1; 4, 2)是否正确，但这是我提出的解决方案：
mapply(function(x, y) which(x %in% y), df, df1)
a b
[1,] 3 2
[2,] 5 4