我有一个反应式数据帧，如下所示：
Apr 2017 May 2017 Jun 2017 Jul 2017 Aug 2017 Sep 2017
zz 0.1937571 0.1840005 0.1807256 0.1959589 0.2039463 0.2016886
aa 0.3518203 0.3634578 0.3670747 0.3676495 0.3680581 0.3657724
bb 0.10651308 0.11548379 0.11572389 0.11272168 0.11361587 0.11503638
cc 0.2481513 0.2579199 0.2623222 0.2673914 0.2579430 0.2550686
dd 0.06641069 0.06741159 0.07305105 0.07373854 0.07043972 0.07304338
我正在尝试根据值(类似于这个，eg3(设置整个表格的样式。 以下是我的代码：
brks <- reactive({
quantile(intrc_pattern_re(), probs = seq(0, 1, 0.25), na.rm = TRUE)
})
clrs <- reactive({
round(seq(255, 40, length.out = length(brks()) + 1), 0) %>%
paste0("rgb(255,", ., ",", ., ")")
})
intrc_pattern_reshape <- reactive ({
datatable(intrc_pattern_re(),
options = list(searching = FALSE,
pageLength = 15,
lengthChange = FALSE)
) %>%
formatPercentage(colnames(intrc_pattern_re()), 2) %>%
formatStyle(names(intrc_pattern_re()),
backgroundColor = styleInterval(brks(), clrs()))
})
但是当我这样做时，我收到以下错误：non-numeric argument to binary operator
有人可以告诉我我做错了什么吗？谢谢。 dput(df，"( 的输出
structure(list(`Apr 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.06641069",
"0.10651308", "0.1937571", "0.2481513", "0.3090870", "0.3518203",
"0.4697810", "Apr 2017"), class = "factor"), `May 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.06741159",
"0.11548379", "0.1840005", "0.2579199", "0.3043959", "0.3634578",
"0.4719425", "May 2017"), class = "factor"), `Jun 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.07305105",
"0.11572389", "0.1807256", "0.2623222", "0.3030102", "0.3670747",
"0.4766237", "Jun 2017"), class = "factor"), `Jul 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.07373854",
"0.11272168", "0.1959589", "0.2673914", "0.2984132", "0.3676495",
"0.4759238", "Jul 2017"), class = "factor"), `Aug 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.07043972",
"0.11361587", "0.2039463", "0.2579430", "0.2970350", "0.3680581",
"0.4828409", "Aug 2017"), class = "factor"), `Sep 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.07304338",
"0.11503638", "0.2016886", "0.2550686", "0.2998945", "0.3657724",
"0.4909182", "Sep 2017"), class = "factor"), `Oct 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 2L, `cc` = 4L,
dd = 1L, Premium = 7L, `ff` = 5L), .Label = c("0.07651393",
"0.11219458", "0.2025043", "0.2479362", "0.2866641", "0.3673334",
"0.5121613", "Oct 2017"), class = "factor"), `Nov 2017` = structure(c(`zz` = 3L,
aa = 6L, `bb` = 1L, `cc` = 4L,
dd = 2L, Premium = 7L, `ff` = 5L), .Label = c("0.10724728",
"0.15016708", "0.1857769", "0.2280702", "0.2691103", "0.3417920",
"0.4948308", "Nov 2017"), class = "factor"), `Dec 2017` = structure(c(`zz` = 2L,
aa = 5L, `bb` = 1L, `cc` = 3L,
dd = 6L, Premium = 7L, `ff` = 4L), .Label = c("0.08775835",
"0.1659323", "0.1945492", "0.2304338", "0.2958437", "0.29888712",
"0.4493300", "Dec 2017"), class = "factor"), `Jan 2018` = structure(c(`zz` = 2L,
aa = 5L, `bb` = 1L, `cc` = 3L,
dd = 6L, Premium = 7L, `ff` = 4L), .Label = c("0.08016616",
"0.1565603", "0.1753247", "0.2134740", "0.2811306", "0.34148205",
"0.4315794", "Jan 2018"), class = "factor")), row.names = c("zz",
"aa", "bb", "cc", "dd",
"Premium", "ff"), class = "data.frame")
您得到的错误：当您将不属于numeric类型的内容传递给二进制运算符(如+或-时，会发生non-numeric argument to binary operator。例如：
> 'a'+3
Error in "a" + 3 : non-numeric argument to binary operator
您在调用quantile时收到此错误，因为您作为intrc_pattern_re()传入的 data.frame 中的所有数字都被错误地分类为factors而不是numeric。如果你看一下dput的输出，你会看到每一行都说class = "factor"))
quantile中的某个地方是一个二进制运算符，它期望接收numeric并在获得factor时抛出错误。
要解决这个问题，你只需要将intrc_pattern_re()返回的data.frame的每一列都变成numeric。
如果我们df加载您的数据框：
quantile(x, probs = seq(0, 1, 0.25), na.rm = TRUE)
Error in (1 - h) * qs[i] : non-numeric argument to binary operator
如果我们将这些factor变量转换为numeric(请注意，您必须首先转换为character然后转换为numeric(，那么它可以工作：
df2 <- df %>%
dplyr::mutate_if(is.factor, function(x) as.numeric(as.character(x)))
quantile(df2, probs = seq(0, 1, 0.25), na.rm = TRUE)
0% 25% 50% 75% 100%
0.06641069 0.15176539 0.25160995 0.34171451 0.51216130