Question

我在网上发现了一个代码，必须（将）生成代表美国劳工部：劳工统计局的一些数据的图形。：

library(ggplot2)
df <- as.data.frame(read.csv("unemp.csv", colClasses = c("Date", "numeric")))
p <- ggplot(df,aes(x=date,y=ratio))
p + geom_point() + geom_smooth() + xlab("Year") + 
ylab("Civilian Employment Population Ratio (%)") + 
labs(title="Bureau of Labor Statistics Series EMRATIO 
            (seasonally adjusted) to 2012-10-01")

但它不起作用并产生此错误：

Don't know how to automatically pick scale for object of type function. Defaulting to continuous
Error in eval(expr, envir, enclos) : object 'ratio' not found

此代码中缺少什么？

'unemp.csv'包含来自here的数据，生成的图片必须看起来像this。

Answer 1

出现错误是因为没有任何名为“date”和“ratio”的变量。这很好用：

 library(ggplot2)
df <- as.data.frame(read.table("unemp.txt", header = TRUE, colClasses = c("Date", "numeric")))
names(df) <- c("date", "ratio")
p <- ggplot(df,aes(x=date,y=ratio))
p + geom_point() + geom_smooth() + xlab("Year") + 
  ylab("Civilian Employment Population Ratio (%)") +
  labs(title="Bureau of Labor Statistics Series EMRATIO (seasonally adjusted) to 2012-10-01")

Answer 2

像https://stackoverflow.com/questions/16201906/how-can-i-get-the-value-of-a-kernel-density-estimate-at-specific-points之类的东西可能对任何寻找热图的人都有用。

ggplot2错误：找不到对象'比率'

2 个答案: