我有一个像这样的矩阵：
df<-data.frame(a=c(1,2,5,4,5,4), b=c(3,4,8,6,7,4))
我想知道以下矩阵是否包含在前一个矩阵中以及在哪里：
df1<-data.frame(a=c(5,4), b=c(7,4))
我知道如何查找元素：
which( df ==df1[1,1], arr.ind=T )
但不是完整的矩阵。我需要获取大矩阵中子矩阵的坐标。在这种情况下将是
(5,1;6,2)
有没有办法不用循环就可以解决这个问题？
【问题讨论】：
df1<-data.frame(a=c(5,4), b=c(8,6)) 吗？
Rcpp 是解决此类问题的好工具。
我在这里有点过火了，写了一个非常复杂的函数，它可以找到一个较大数组中较小数组的所有匹配项的最低索引（矩阵的左上角）角的坐标，对于任何维度。如果你想在一个11维数组中找到一个9维数组的所有位置，这个函数可以帮你搞定。
这里是：
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;
}
');
这是我做的一些相当随意的测试，最多只是立方体中的立方体（每个测试打印 big 数组，然后是 small 数组，然后是结果，最后是逻辑向量测试，如果切片从big 中的每个连续匹配延伸的small 的大小实际上与small 相同）：
## 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
这里有一个关于你的数据的演示：
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
我的函数只返回索引最低的坐标，因为你可以通过简单地加上small的大小来得出最高索引的坐标，如下：
t(t(findarray(as.matrix(df),as.matrix(df1)))+dim(df1))-1;
## [,1] [,2]
## [1,] 6 2
请注意，转置是必要的，因为 R 循环短向量对更大矩阵的方式（即跨行，然后跨列）。对于您的特定数据，这显然不是必需的，因为只有一个匹配项，而且df1 的两个维度具有相同的长度，所以无论如何都没关系，但在一般情况下很重要。
好吧，我可以说我做到了，这是一个在 11D 数组中匹配 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
想到了另一种测试的好方法：我们可以从大数组中取出切片，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
【讨论】：
说实话，我真的不认为有办法避免循环：
# 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
【讨论】：
我不知道你的答案(3, 1; 4, 2)是否正确，但这是我想出的解决方案：
mapply(function(x, y) which(x %in% y), df, df1)
a b
[1,] 3 2
[2,] 5 4
【讨论】：