我有一个函数,它使用ggplot2获取数据帧并从该数据帧中绘制多个列。 ggplot2中的aes()函数采用label参数,我想使用sprintf格式化该参数 - 这是我之前在其他代码中做过很多次的事情。当我将格式字符串传递给sprintf(在本例中为“%1.1f”)时,它表示“找不到对象”。如果我使用round()函数并将参数传递给该函数,它可以毫无问题地找到它。格式()也是如此。显然只有sprintf()无法看到对象。
起初我认为这是一个懒惰的评估问题,因为调用函数而不是使用内联代码,但在我传递给sprintf的格式字符串上使用force()并不能解决问题。我可以解决这个问题,但我想知道为什么会这样。当然,我可能忽略了一些微不足道的事情。
Q值。为什么sprintf()找不到字符串对象?
代码如下(编辑和修剪以获得更少的示例)
require(gdata)
require(ggplot2)
require(scales)
require(gridExtra)
require(lubridate)
require(plyr)
require(reshape)
set.seed(12345)
# Create dummy time series data with year and month
monthsback <- 64
startdate <- as.Date(paste(year(now()),month(now()),"1",sep = "-")) - months(monthsback)
mydf <- data.frame(mydate = seq(as.Date(startdate), by = "month", length.out = monthsback), myvalue5 = runif(monthsback, min = 200, max = 300))
mydf$year <- as.numeric(format(as.Date(mydf$mydate), format="%Y"))
mydf$month <- as.numeric(format(as.Date(mydf$mydate), format="%m"))
getchart_highlight_value <- function(
plotdf,
digits_used = 1
)
{
force(digits_used)
#p <- ggplot(data = plotdf, aes(x = month(mydate, label = TRUE), y = year(mydate), fill = myvalue5, label = round(myvalue5, digits_used))) +
# note that the line below using sprintf() does not work, whereas the line above using round() is fine
p <- ggplot(data = plotdf, aes(x = month(mydate, label = TRUE), y = year(mydate), fill = myvalue5, label = sprintf(paste("%1.",digits_used,"f", sep = ""), myvalue5))) +
scale_x_date(labels = date_format("%Y"), breaks = date_breaks("years")) +
scale_y_reverse(breaks = 2007:2012, labels = 2007:2012, expand = c(0,0)) +
geom_tile() + geom_text(size = 4, colour = "black") +
scale_fill_gradient2(low = "blue", high = "red", limits = c(min(plotdf$myvalue5), max(plotdf$myvalue5)), midpoint = median(plotdf$myvalue5)) +
scale_x_discrete(expand = c(0,0)) +
opts(panel.grid.major = theme_blank()) +
opts(panel.background = theme_rect(fill = "transparent", colour = NA)) +
png(filename = "c:/sprintf_test.png", width = 700, height = 300, units = "px", res = NA)
print(p)
dev.off()
}
getchart_highlight_value (plotdf <- mydf,
digits_used <- 1)
答案 0 :(得分:4)
这是一个最小的例子
require(ggplot2)
getchart_highlight_value <- function(df)
{
fmt <- "%1.1f"
ggplot(df, aes(x, x, label=sprintf(fmt, lbl))) + geom_tile()
}
df <- data.frame(x = 1:5, lbl = runif(5))
getchart_highlight_value (df)
失败了
> getchart_highlight_value (df)
Error in sprintf(fmt, lbl) : object 'fmt' not found
如果我在全球环境中创建fmt
那么一切都很好;也许这解释了'有时它有用'/'它对我有用'以上评论。
> sessionInfo()
R version 2.15.0 Patched (2012-05-01 r59304)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=C LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] ggplot2_0.9.1
loaded via a namespace (and not attached):
[1] colorspace_1.1-1 dichromat_1.2-4 digest_0.5.2 grid_2.15.0
[5] labeling_0.1 MASS_7.3-18 memoise_0.1 munsell_0.3
[9] plyr_1.7.1 proto_0.3-9.2 RColorBrewer_1.0-5 reshape2_1.2.1
[13] scales_0.2.1 stringr_0.6
答案 1 :(得分:4)
使用Martin的最小示例(这是一个最小示例,另请参阅this question),您可以通过指定ggplot()
应该使用的环境来使代码工作。为此,请在environment
函数中指定参数ggplot()
,例如:
require(ggplot2)
getchart_highlight_value <- function(df)
{
fmt <- "%1.1f"
ggplot(df, aes(x, x, label=sprintf(fmt, lbl)),
environment = environment()) +
geom_tile(bg="white") +
geom_text(size = 4, colour = "black")
}
df <- data.frame(x = 1:5, lbl = runif(5))
getchart_highlight_value (df)
函数environment()
返回当前(本地)环境,该环境是函数getchart_highlight_value()
创建的环境。如果您未指定此内容,ggplot()
将查看全局环境,并且未定义变量fmt
。
与懒惰评估无关,与选择合适的环境无关。
上面的代码产生了以下情节: