从R格式化JSON输出的策略
从R格式化JSON输出的策略
提问于 2011-11-28 04:50:59
我正在尝试找出从R生成JSON文件的最佳方法。我在R中有以下数据帧tmp。
> tmp
gender age welcoming proud tidy unique
1 1 30 4 4 4 4
2 2 34 4 2 4 4
3 1 34 5 3 4 5
4 2 33 2 3 2 4
5 2 28 4 3 4 4
6 2 26 3 2 4 3dput(tmp)的输出如下：
tmp <- structure(list(gender = c(1L, 2L, 1L, 2L, 2L, 2L), age = c(30,
34, 34, 33, 28, 26), welcoming = c(4L, 4L, 5L, 2L, 4L, 3L), proud = c(4L,
2L, 3L, 3L, 3L, 2L), tidy = c(4L, 4L, 4L, 2L, 4L, 4L), unique = c(4L,
4L, 5L, 4L, 4L, 3L)), .Names = c("gender", "age", "welcoming",
"proud", "tidy", "unique"), na.action = structure(c(15L, 39L,
60L, 77L, 88L, 128L, 132L, 172L, 272L, 304L, 305L, 317L, 328L,
409L, 447L, 512L, 527L, 605L, 618L, 657L, 665L, 670L, 708L, 709L,
729L, 746L, 795L, 803L, 826L, 855L, 898L, 911L, 957L, 967L, 983L,
984L, 988L, 1006L, 1161L, 1162L, 1224L, 1245L, 1256L, 1257L,
1307L, 1374L, 1379L, 1386L, 1387L, 1394L, 1401L, 1408L, 1434L,
1446L, 1509L, 1556L, 1650L, 1717L, 1760L, 1782L, 1814L, 1847L,
1863L, 1909L, 1930L, 1971L, 2004L, 2022L, 2055L, 2060L, 2065L,
2082L, 2109L, 2121L, 2145L, 2158L, 2159L, 2226L, 2227L, 2281L
), .Names = c("15", "39", "60", "77", "88", "128", "132", "172",
"272", "304", "305", "317", "328", "409", "447", "512", "527",
"605", "618", "657", "665", "670", "708", "709", "729", "746",
"795", "803", "826", "855", "898", "911", "957", "967", "983",
"984", "988", "1006", "1161", "1162", "1224", "1245", "1256",
"1257", "1307", "1374", "1379", "1386", "1387", "1394", "1401",
"1408", "1434", "1446", "1509", "1556", "1650", "1717", "1760",
"1782", "1814", "1847", "1863", "1909", "1930", "1971", "2004",
"2022", "2055", "2060", "2065", "2082", "2109", "2121", "2145",
"2158", "2159", "2226", "2227", "2281"), class = "omit"), row.names = c(NA,
6L), class = "data.frame")使用rjson包，我运行toJSON(tmp)行，它将生成以下JSON文件：
{"gender":[1,2,1,2,2,2],
"age":[30,34,34,33,28,26],
"welcoming":[4,4,5,2,4,3],
"proud":[4,2,3,3,3,2],
"tidy":[4,4,4,2,4,4],
"unique":[4,4,5,4,4,3]}我还试验了RJSONIO包；toJSON()的输出是相同的。我想要生成的是以下结构：
{"traits":["gender","age","welcoming","proud", "tidy", "unique"],
"values":[
{"gender":1,"age":30,"welcoming":4,"proud":4,"tidy":4, "unique":4},
{"gender":2,"age":34,"welcoming":4,"proud":2,"tidy":4, "unique":4},
....
]我不知道怎么做才是最好的。我意识到我可以使用python逐行解析它，但我觉得可能有更好的方法来做到这一点。我还意识到R中的数据结构没有反映JSON文件(特别是traits行)中所需的元信息，但我主要感兴趣的是生成像该行这样格式化的数据
{"gender":1,"age":30,"welcoming":4,"proud":4,"tidy":4, "unique":4}因为我可以手动添加第一行。
编辑:我发现了一个有用的blog帖子，其中作者处理了类似的问题并提供了解决方案。此函数从数据帧生成格式化的JSON文件。
toJSONarray <- function(dtf){
clnms <- colnames(dtf)
name.value <- function(i){
quote <- '';
# if(class(dtf[, i])!='numeric'){
if(class(dtf[, i])!='numeric' && class(dtf[, i])!= 'integer'){ # I modified this line so integers are also not enclosed in quotes
quote <- '"';
}
paste('"', i, '" : ', quote, dtf[,i], quote, sep='')
}
objs <- apply(sapply(clnms, name.value), 1, function(x){paste(x, collapse=', ')})
objs <- paste('{', objs, '}')
# res <- paste('[', paste(objs, collapse=', '), ']')
res <- paste('[', paste(objs, collapse=',\n'), ']') # added newline for formatting output
return(res)
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/8288925
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