我写了一个简短的脚本,用于绘制2个独立测量设备上测量的放射性活动的趋势。该脚本如下所示
pkgLoad <- function(x)
{
if (!require(x,character.only = TRUE))
{
install.packages(x,dep=TRUE, repos='http://star-www.st-andrews.ac.uk/cran/')
if(!require(x,character.only = TRUE)) stop("Package not found")
}
}
pkgLoad("ggplot2")
pkgLoad("XLConnect")
pkgLoad("reshape2")
#Load the workbook
wb<-loadWorkbook("CapintecQC.xlsx")
df_blue <-readWorksheet(wb, sheet = "Blue", startCol=1, endCol=6)
#sort date format
df_blue$Date <- as.Date(df_blue$Date , "%d/%m/%y")
df_blue[order(df_blue$Date ),]
df_gold <-readWorksheet(wb, sheet = "Gold", startCol=1, endCol=6)
df_gold$Date <- as.Date(df_gold$Date , "%d/%m/%y")
df_gold[order(df_gold$Date ),]
#Reference Cs-137 details
ref_activity <- 9.3
half_life <- 30.23
ref_date <- as.Date('06/01/08',format='%d/%m/%y')
blue_melt <- melt(df_blue[,c(1,2:6)], id="Date", value.name="Activity", variable.name="Isotope")
#Add new column to data frame with expected activity
df_gold["Exp_Act"] <- round(ref_activity*exp((-0.693/half_life)*as.numeric(difftime(df_gold$Date,ref_date))/365.25),3)
df_gold["Exp_Act_0.95"] <- 0.95 * df_gold$Exp_Act
df_gold["Exp_Act_1.05"] <- 1.05 * df_gold$Exp_Act
gold_melt <- melt(df_gold[,c(1,2:6)], id="Date", value.name="Activity", variable.name="Isotope")
p <- ggplot( NULL )+geom_point(data = gold_melt, aes(x=Date,y=Activity, col=Isotope)) + geom_ribbon(data = df_gold, aes(x = Date, ymin = Exp_Act_0.95, ymax = Exp_Act_1.05), fill='blue', alpha=0.2) + geom_point(data = blue_melt, aes(x=Date,y=Activity, col=Isotope), shape=2) + theme_bw()
print(p)
我对R / ggplot2不太称职。我想最后的图表显示每种放射性核素的测量活性对于两种器件都是相同的颜色(即红色的Cs-137,蓝色的99mTc)。我怎么能这样做,因为我的图表绘制了不同的颜色。
这个传说也不令人满意。 (i)从excel标题中拾取的每个核素的格式从Cs-137变为Cs.137。如何将Cs-137,Tc-99m等作为标题? (ii)每个放射性核素都在图例中重复 - 每个设备一个。是否可以只显示第一个数据框(df_gold)的图例,或者更好的只是在图例中有文字,文字颜色与图中的标记颜色相匹配?)
df_gold结构
structure(list(Date = structure(c(15708, 15709, 15712, 15713,
15714, 15715, 15716, 15719, 15720, 15721, 15722, 15723, 15726,
15727, 15729, 15730, 15733, 15734, 15735, 15736, 15740, 15741,
15743, 15747, 15748, 15749, 15750, 15751, 15754, 15755, 15756,
15757, 15758, 15761, 15762, 15764, 15765, 15768, 15769, 15770,
15771, 15772, 15775, 15776, 15777, 15779, 15782, 15783, 15784,
15785, 15786, 15789, 15790, 15791, 15792, 15797, 15798, 15799,
15800), class = "Date"), Cs..137 = c(8.2, 8.1, 8.1, 8.1, 8.1,
8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1,
8.1, 8.1, 8.1, 8.2, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8, 8.2,
8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1,
8.1, 8.1, 8.1, 8, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1, 8.1,
8.1, 8.1), In..111 = c(6.49, 6.47, 6.48, 6.43, 6.49, 6.51, 6.5,
6.47, 6.48, 6.4, 6.48, 6.48, 6.48, 6.49, 6.49, 6.47, 6.48, 6.48,
6.5, 6.47, 6.49, 6.55, 6.46, 6.49, 6.48, 6.48, 6.46, 6.48, 6.49,
6.44, 6.49, 6.46, 6.45, 6.46, 6.46, 6.43, 6.49, 6.47, 6.45, 6.43,
6.44, 6.44, 6.44, 6.46, 6.45, 6.47, 6.45, 6.43, 6.44, 6.47, 6.45,
6.46, 6.45, 6.46, 6.39, 6.46, 6.44, 6.42, 6.41), I..123 = c(6.97,
6.94, 6.96, 6.91, 6.92, 6.95, 6.93, 6.92, 6.93, 7, 6.97, 6.96,
6.96, 6.94, 6.98, 6.97, 6.95, 6.95, 6.94, 6.96, 6.97, 7.01, 6.92,
7, 6.98, 6.97, 6.91, 6.99, 6.95, 6.88, 6.96, 6.91, 6.91, 6.93,
6.94, 6.94, 6.97, 6.93, 6.93, 6.93, 6.96, 6.94, 6.94, 6.92, 6.93,
6.91, 6.93, 6.92, 6.92, 6.91, 6.91, 6.89, 6.92, 6.9, 6.9, 6.91,
6.91, 6.9, 6.9), I..131 = c(10.5, 10.5, 10.5, 10.5, 10.5, 10.5,
10.5, 10.5, 10.5, 10.8, 10.5, 10.6, 10.5, 10.5, 10.5, 10.5, 10.5,
10.5, 10.5, 10.5, 10.5, 10.6, 10.5, 10.5, 10.5, 10.5, 10.5, 10.5,
10.5, 10.5, 10.5, 10.5, 10.5, 10.5, 10.5, 10.5, 10.5, 10.5, 10.4,
10.5, 10.4, 10.5, 10.5, 10.5, 10.4, 10.5, 10.4, 10.4, 10.5, 10.4,
10.4, 10.4, 10.4, 10.4, 10.3, 10.5, 10.5, 10.5, 10.6), Tc..99m = c(15,
15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15.1, 15, 15, 15.1, 15,
15, 15, 15, 15.1, 15, 15.1, 15, 15, 15, 15, 15, 15, 15, 15, 15,
15, 15, 15, 15, 14.9, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15,
14.9, 14.8, 14.9, 14.9, 14.9, 14.9, 15, 15, 14.8, 15, 15, 15,
15), Exp_Act = c(8.294, 8.293, 8.292, 8.291, 8.291, 8.29, 8.29,
8.288, 8.288, 8.287, 8.287, 8.286, 8.285, 8.284, 8.283, 8.283,
8.281, 8.28, 8.28, 8.279, 8.277, 8.277, 8.276, 8.274, 8.273,
8.273, 8.272, 8.272, 8.27, 8.27, 8.269, 8.269, 8.268, 8.266,
8.266, 8.265, 8.264, 8.263, 8.262, 8.262, 8.261, 8.261, 8.259,
8.259, 8.258, 8.257, 8.256, 8.255, 8.255, 8.254, 8.254, 8.252,
8.251, 8.251, 8.25, 8.248, 8.247, 8.247, 8.246), Exp_Act_0.95 = c(7.8793,
7.87835, 7.8774, 7.87645, 7.87645, 7.8755, 7.8755, 7.8736, 7.8736,
7.87265, 7.87265, 7.8717, 7.87075, 7.8698, 7.86885, 7.86885,
7.86695, 7.866, 7.866, 7.86505, 7.86315, 7.86315, 7.8622, 7.8603,
7.85935, 7.85935, 7.8584, 7.8584, 7.8565, 7.8565, 7.85555, 7.85555,
7.8546, 7.8527, 7.8527, 7.85175, 7.8508, 7.84985, 7.8489, 7.8489,
7.84795, 7.84795, 7.84605, 7.84605, 7.8451, 7.84415, 7.8432,
7.84225, 7.84225, 7.8413, 7.8413, 7.8394, 7.83845, 7.83845, 7.8375,
7.8356, 7.83465, 7.83465, 7.8337), Exp_Act_1.05 = c(8.7087, 8.70765,
8.7066, 8.70555, 8.70555, 8.7045, 8.7045, 8.7024, 8.7024, 8.70135,
8.70135, 8.7003, 8.69925, 8.6982, 8.69715, 8.69715, 8.69505,
8.694, 8.694, 8.69295, 8.69085, 8.69085, 8.6898, 8.6877, 8.68665,
8.68665, 8.6856, 8.6856, 8.6835, 8.6835, 8.68245, 8.68245, 8.6814,
8.6793, 8.6793, 8.67825, 8.6772, 8.67615, 8.6751, 8.6751, 8.67405,
8.67405, 8.67195, 8.67195, 8.6709, 8.66985, 8.6688, 8.66775,
8.66775, 8.6667, 8.6667, 8.6646, 8.66355, 8.66355, 8.6625, 8.6604,
8.65935, 8.65935, 8.6583)), row.names = c(NA, -59L), .Names = c("Date",
"Cs..137", "In..111", "I..123", "I..131", "Tc..99m", "Exp_Act",
"Exp_Act_0.95", "Exp_Act_1.05"), class = "data.frame")
df_blue结构
structure(list(Date = structure(c(15790, 15791, 15792, 15797,
15798, 15799, 15800), class = "Date"), Cs.137 = c(8.1, 8.2, 8.2,
8.2, 8.2, 8.2, 8.2), I.123 = c(6.82, 6.85, 6.91, 6.84, 6.82,
6.82, 6.83), I.131 = c(10.5, 10.6, 10.6, 10.5, 10.6, 10.6, 10.6
), In.111 = c(6.35, 6.45, 6.43, 6.37, 6.38, 6.4, 6.37), X99m.Tc = c(15,
15, 15.1, 15.1, 15.1, 15.1, 15.1)), .Names = c("Date", "Cs.137",
"I.123", "I.131", "In.111", "X99m.Tc"), row.names = c(NA, -7L
), class = "data.frame")
答案 0 :(得分:1)
我的方法是将两个数据框绑定在一起,然后添加包含设备名称(金色或蓝色)的新列。
df<-rbind(gold_melt,blue_melt)
df$device<-rep(c("gold","blue"),c(nrow(gold_melt),nrow(blue_melt)))
使用库recode()
中的函数car
更改Isotope
的名称。
df$Isotope<-recode(df$Isotope,"c('Cs..137','Cs.137')='Cs-137';
c('I..123','I.123')='I-123';
c('I..131','I.131')='I-131';
c('In..111','In.111')='In-111'
;c('Tc..99m','X99m.Tc')='Tc-99m'")
现在您只需要使用新数据框拨打geom_point()
。我还添加了shape=device
以获得每个设备的不同形状。
ggplot(NULL) +
geom_point(data=df,aes(x=Date,y=Activity, col=Isotope,shape=device))+
geom_ribbon(data = df_gold, aes(x = Date, ymin = Exp_Act_0.95, ymax = Exp_Act_1.05), fill='blue', alpha=0.2)
答案 1 :(得分:1)
万一你想要将这些传说融合在一起,然后建立在Didzis的答案上:
df <- transform(df, device = factor(device, levels=unique(device)),
grp = paste(Isotope, device, sep="_"))
require(RColorBrewer)
ggplot() + geom_point(data = df, aes(x = Date, y = Activity,
colour=grp, shape = grp, fill=grp)) +
geom_ribbon(data = df_gold, aes(x = Date, ymin = Exp_Act_0.95,
ymax = Exp_Act_1.05), fill='blue', alpha=0.2) +
scale_shape_manual("", values=rep(c(21,24), 5)) +
scale_fill_manual("", values=rep(brewer.pal(5, "Set1"), each=2)) +
scale_colour_manual("", values=rep(brewer.pal(5, "Set1"), each=2))