我有一些图像数据作为bytea存储在PostgreSQL数据库表列中。我也有关于数据的元数据,可用于解释它,相关的是图像尺寸和类。类包括int16,uint16。我找不到有关在R中正确解释带符号/无符号整数的任何信息。
我正在使用RPostgreSQL将数据提取到R中,我想在R中查看图像。
MWE:
# fakeDataQuery <- dbGetQuery(conn,
# 'select byteArray, ImageSize, ImageClass from table where id = 1')
# Example 1 (no negative numbers)
# the actual byte array shown in octal sequences in pgadmin (1.22.2) Query Output is:
# "\001\000\002\000\003\000\004\000\005\000\006\000\007\000\010\000\011\000"
# but RPostgreSQL returns the hex-encoded version:
byteArray <- "\\x010002000300040005000600070008000900"
ImageSize <- c(3, 3, 1)
ImageClass <- 'int16'
# expected result
> array(c(1,2,3,4,5,6,7,8,9), dim=c(3,3,1))
# , , 1
#
# [,1] [,2] [,3]
#[1,] 1 4 7
#[2,] 2 5 8
#[3,] 3 6 9
# Example 2: (with negtive numbers)
byteArray <- "\\xffff00000100020003000400050006000700080009000a00"
ImageSize <- c(3, 4, 1)
ImageClass <- 'int16'
# expectedResult
> array(c(-1,0,1,2,3,4,5,6,7,8,9,10), dim=c(3,4,1))
#, , 1
#
# [,1] [,2] [,3] [,4]
#[1,] -1 2 5 8
#[2,] 0 3 6 9
#[3,] 1 4 7 10
我尝试过的:
来自PostgreSQL的BYtea数据是一个长字符串,编码为“十六进制”,可以通过前面的\\x
来分辨(我相信还有一个额外的\
转义现有代码?):https://www.postgresql.org/docs/9.1/static/datatype-binary.html(请参阅:第8.4.1节“ bytea十六进制格式”)
将'hex'解码回原始类型(基于ImageClass的'int16')
在上面的the same url中,十六进制编码使用“每个字节2个十六进制数字”。因此,我需要将编码的byteArray拆分为适当长度的子字符串,请参见:this link
# remove the \\x hex encoding indicator(s) added by PostgreSQL
byteArray <- gsub("\\x", "", x = byteArray, fixed=T)
l <- 2 # hex digits per byte (substring length)
byteArray <- strsplit(trimws(gsub(pattern = paste0("(.{",l,"})"),
replacement = "\\1 ",
x = byteArray)),
" ")[[1]]
# for some reason these appear to be in the opposite order than i expect
# Ex: 1 is stored as '0100' rather than '0001'
# so reverse the digits (int16 specific)
byteArray <- paste0(byteArray[c(F,T)],byteArray[c(T,F)])
# strtoi() converts a vector of hex values given a decimal base
byteArray <- strtoi(byteArray, 16L)
# now make it into an n x m x s array,
# e.g., 512 x 512 x (# slices)
V = array(byteArray, dim = ImageSize)
此解决方案有两个问题:
任何人都可以使用签名类型的解决方案吗?
答案 0 :(得分:1)
您可以从this conversion function开始,用更快的strsplit
代替,然后在结果上使用readBin
:
byteArray <- "\\xffff00000100020003000400050006000700080009000a00"
## Split a long string into a a vector of character pairs
Rcpp::cppFunction( code = '
CharacterVector strsplit2(const std::string& hex) {
unsigned int length = hex.length()/2;
CharacterVector res(length);
for (unsigned int i = 0; i < length; ++i) {
res(i) = hex.substr(2*i, 2);
}
return res;
}')
## A function to convert one string to an array of raw
f <- function(x) {
## Split a long string into a a vector of character pairs
x <- strsplit2(x)
## Remove the first element, "\\x"
x <- x[-1]
## Complete the conversion
as.raw(as.hexmode(x))
}
raw <- f(byteArray)
# int16
readBin(con = raw,
what = "integer",
n = length(raw) / 2,
size = 2,
signed = TRUE,
endian = "little")
# -1 0 1 2 3 4 5 6 7 8 9 10
# uint16
readBin(con = raw,
what = "integer",
n = length(raw) / 2,
size = 2,
signed = FALSE,
endian = "little")
# 65535 0 1 2 3 4 5 6 7 8 9 10
# int32
readBin(con = raw,
what = "integer",
n = length(raw) / 4,
size = 4,
signed = TRUE,
endian = "little")
# 65535 131073 262147 393221 524295 655369
但这不适用于uint32
和(u)int64
,因为R在内部使用int32
。但是,R也可以使用numerics
存储2 ^ 52以下的整数。这样我们就可以使用:
# uint32
byteArray <- "\\xffffffff0100020003000400050006000700080009000a00"
int32 <- readBin(con = f(byteArray),
what = "integer",
n = length(raw) / 4,
size = 4,
signed = TRUE,
endian = "little")
ifelse(int32 < 0, int32 + 2^32, int32)
# 4294967295 131073 262147 393221 524295 655369
对于gzip
压缩数据:
# gzip
byteArray <- "\\x1f8b080000000000000005c1870100200800209a56faffbd41d30dd3b285e37a52f9d033018818000000"
con <- gzcon(rawConnection(f(byteArray)))
readBin(con = con,
what = "integer",
n = length(raw) / 2,
size = 2,
signed = TRUE,
endian = "little")
close(con = con)
由于这是真正的连接,所以我们必须确保将其关闭。