2

frank我对这个功能感到困惑。这里的文档说:

仅适用于列表、data.frames 和 data.tables。计算排名所依据的列。不要引用列名。如果 ... 缺失,则默认考虑所有列。要按列降序排序前缀“-”,例如 frank(x, a, -b, c)。当 b 也是字符类型时, -b 也有效。

所以我有我的数据:

structure(list(product = c("Product 1", "Product 1", "Product 1", 
                           "Product 1", "Product 1", "Product 5", "Product 5", "Product 5", 
                           "Product 5", "Product 5"), policyID = c("A738-33", "A738-33", 
                                                                   "A738-33", "A738-33", "A738-33", "A738-33", "A738-33", 
                                                                   "A738-33", "A738-33", "A738-33"), startYear = c(2014, 
                                                                                                                               2015, 2016, 2017, 2018, 2014, 2015, 2016, 2017, 2018), total = c("30000", 
                                                                                                                                                                                                     "30000", "30000", "30000", "30000", "10000", "10000", "10000", 
                                                                                                                                                                                                     "10000", "10000"), daily = c("150", "150", "150", "150", "150", 
                                                                                                                                                                                                                                     "80", "80", "80", "80", "80")), class = c("data.table", "data.frame"
                                                                                                                                                                                                                                     ), row.names = c(NA, -10L), .internal.selfref = <pointer: 0x7feec50126e0>, sorted = "product")

我想按列totaldaily. 所以我这样做了:

> setDT(testDT)
> frankv(testDT, totallimit, rbddaily, ties.method="dense")
Error in colnamesInt(x, cols, check_dups = TRUE) : 
  argument specifying columns specify non existing column(s): cols[1]='30000'

奇怪的是,当我使用引号时,与文档所说的完全相反,我得到了结果:

frankv(testDT, cols=c("totallimit", "rbddaily"), ties.method="dense")

我也尝试将thin集成到data.table中,结果又发生了一件奇怪的事情。从我拥有的 10 行数据中,我获得了 100 行。

testDT[,.(rank = frankv(testDT, cols=c("limit", "daily"), ties.method="dense")), by = c("policyID", "product", "startYear")]

我做错了什么,我该如何解决?该文档没有太大帮助,也许我遗漏了一些东西......

4

1 回答 1

1

因为frank你不应该引用,但对于frankv(你使用的函数)你应该:

library(data.table)
frank(testDT, total, daily, ties.method="dense")

 [1] 2 2 2 2 2 1 1 1 1 1

frankv(testDT, cols=c("total", "daily"), ties.method="dense")

 [1] 2 2 2 2 2 1 1 1 1 1
于 2020-06-23T10:41:28.050 回答