R 二进制/十进制转换混淆 - AIS 数据

Question

我正在使用 AIS（自动识别系统）数据来定位船只。我能够按照this guide 成功解码几乎所有信息位（与here 完成的在线解码相比）。

但是，我遇到了经度部分的问题。我认为这与十进制值为负有关，但我不知道要在我的代码中更改什么以使其正确。

TLDR 版本：如何从二进制字符串 1101001000001001001110010000 得到十进制值 -48196720（或 48196720）？

完整版：

玩具数据：

library(dplyr)
library(tidyr)
library(stringr)

# choose an example - two strings are provided.  
# The first string shows the issue with the longitude, 
# whereas the second string (where longitude is positive) has no issue
s <- "15E3tB001;J@BLPaK5j7qFA406;d" 
# s <- "133m@ogP00PD;88MD5MTDww@2D7k"
### for input into the online decoder - use these: 
# !AIVDM,1,1,,A,14eG;o@034o8sd<L9i:a;WF>062D,0*7D
# !AIVDM,1,1,,A,133m@ogP00PD;88MD5MTDww@2D7k,0*46

temp <- data.frame(V6 = s) %>%
    # splitting the AIS info into separate characters
    mutate(char_split = str_split(V6,pattern=""))
temp$Text <- apply(temp, 1, function(x) paste(unlist(x$char_split), collapse = ","))        

temp <- temp %>%
    select(-char_split) %>% 
    # and then into separate columns    
    separate(Text, into = paste0("v", 1:43, sep = ""), sep = ",", fill = "right") 

ASCII <- temp %>%
    select(v1:v43)
# translating to ASCII
ASCII <- apply(ASCII, c(1, 2), function(x) utf8ToInt(x))
# translating to 6-bit
ASCII <- apply(ASCII, c(1, 2), function(x) ifelse(x <= 88, x - 48, x - 48 - 8))

一旦数据是ASCII，需要转换成二进制

# making binary
Binary <- apply(ASCII, c(1, 2), function(x){ paste(rev(as.integer(intToBits(x))[1:6]), collapse = "")})
# pasting all the binary info into a single string
temp$Binary <- apply(Binary, 1, function(x) paste(x, collapse = "" ))
temp <- temp %>%
    select(V6, Binary) %>%
    # selecting bits of the single binary string, 
    #and translating back to decimal, as per the guide
    mutate(MMSI = strtoi(substr(Binary, 9, 38), base = 2),
            SOG = strtoi(substr(Binary, 50, 60), base = 2)/10,
            Accuracy = strtoi(substr(Binary, 61, 61), base = 2),
            Lon = strtoi(substr(Binary, 62, 89), base = 2)/600000,
            Lat = strtoi(substr(Binary, 90, 116), base = 2)/600000,
            COG = strtoi(substr(Binary, 117, 128), base = 2)/10,
            Heading = strtoi(substr(Binary, 129, 137), base = 2))

输出：

select(temp, -Binary, -V6)

当我与在线解码器进行比较时，一切都匹配，除了经度。在解码器中，结果是 80.3278667（尽管它实际上是 -80.3278667），而我的是 367.0646。试图对此进行逆向工程，我查看了temp$Binary 的相关子字符串：

mine <- substr(temp$Binary, 62, 89)
RevEng <- -80.3278667 * 600000
binaryLogic:::as.binary(as.integer(RevEng), signed = FALSE) 
mine

所以看起来 RevEng 值与我的二进制字符串的右尾匹配，但我无法弄清楚为什么它与完整的二进制字符串不匹配，或者从这里做什么......

Answer 1

与博文所说的相反，经度是一个有符号整数。但是，它仅使用 28 位，而 R 在内部使用 32 位。因此，您必须自己处理two's compliment conversion。对于设置了最高位的任何数字，您必须减去2^28，例如：

mine <- "1101001000001001001110010000"
strtoi(mine, base = 2) - 2^28
#> [1] -48196720

您可以使用二进制字符串上的substr 或通过查找数字>= 2^27 来识别这些数字。

顺便说一句，这同样适用于纬度，但修改后它只使用 27 位。