1

我在下面使用 Dygraphdygraphs package

library(dygraphs)
library(htmlwidgets)
library(zoo)

valueFormatter = "function formatValue(v) {
                                            var suffixes = ['', 'K', 'M', 'G', 'T'];
                                            if (v < 1000) return v;
                                            var magnitude = Math.ceil(String(Math.floor(v)).length / 4-1);
                                            if (magnitude > suffixes.length - 1)
                                            magnitude = suffixes.length - 1;
                                            return String(Math.round(v / Math.pow(10, magnitude * 3), 2)) +suffixes[magnitude]
                                        }" 
Data = zoo(matrix(c(10000, 1000000, 100000000, 50000), nc = 1), as.Date(c('2015-01-05', '2016-01-05', '2017-01-05', '2018-01-05'))); colnames(Data) = 'x'
dygraph(Data, main = "") %>% dySeries(c("x")) %>% 
                            dyAxis("y", axisLabelFormatter = JS(valueFormatter), 
                                        valueFormatter     = JS(valueFormatter),
                                        rangePad = 20)

但是,在 Y 轴的标签中,我想为刻度值带来千位分隔符,例如,而不是30000K我想要3,0000K. 有什么办法可以做到这一点。

4

1 回答 1

1

将您的数字转换30000K为千位分隔值30,000K

正则表达式在每个第三位数字后放置逗号

var regex = /\B(?=(\d{3})+(?!\d))/g;
var number = "200";
var thousandSeprator = "20000";
console.log(number.replace(regex,',')); // 200
console.log(thousandSeprator.replace(regex,',')); // 20,000

更新valueFormatter功能

valueFormatter = "function formatValue(v) {
                                            var suffixes = ['', 'K', 'M', 'G', 'T'];
                                            var regex = /\B(?=(\d{3})+(?!\d))/g; 
                                            if (v < 1000) return v.toString().replace(regex,',');
                                            var magnitude = Math.ceil(String(Math.floor(v)).length / 4-1);
                                            if (magnitude > suffixes.length - 1)
                                            magnitude = suffixes.length - 1;        
                                            return v.toString().replace(regex,',') +suffixes[magnitude]
                                        }"
于 2018-11-10T07:44:17.320 回答