【发布时间】:2016-04-29 05:56:27
【问题描述】:
我快到了,只是缺少垂直绘制的各个点。
这里是数据:
q6 <- structure(list(x1 = c(0.0629, 0.063, 0.0628, 0.0634, 0.0619,
0.0613, 0.063, 0.0628, 0.0623, 0.0631, 0.0635, 0.0623, 0.0635,
0.0645, 0.0619, 0.0631, 0.0616, 0.063, 0.0636, 0.064, 0.0628,
0.0615, 0.063, 0.0635, 0.0623), x2 = c(0.0636, 0.0631, 0.0631,
0.063, 0.0628, 0.0629, 0.0639, 0.0627, 0.0626, 0.0631, 0.063,
0.063, 0.0631, 0.064, 0.0644, 0.0627, 0.0623, 0.063, 0.0631,
0.0635, 0.0625, 0.0625, 0.0632, 0.0629, 0.0629), x3 = c(0.064,
0.0622, 0.0633, 0.0631, 0.063, 0.0634, 0.0625, 0.0622, 0.0633,
0.0633, 0.0638, 0.063, 0.063, 0.0631, 0.0632, 0.063, 0.0631,
0.0626, 0.0629, 0.0629, 0.0616, 0.0619, 0.063, 0.0635, 0.063),
x4 = c(0.0635, 0.0625, 0.0633, 0.0632, 0.0619, 0.0625, 0.0629,
0.0625, 0.063, 0.0631, 0.0635, 0.0627, 0.063, 0.064, 0.0622,
0.0628, 0.062, 0.0629, 0.0635, 0.0635, 0.062, 0.0619, 0.0631,
0.0631, 0.0626), x5 = c(0.064, 0.0627, 0.063, 0.0633, 0.0625,
0.0628, 0.0627, 0.0627, 0.0624, 0.063, 0.0633, 0.0629, 0.063,
0.0642, 0.0635, 0.0629, 0.0625, 0.0628, 0.0634, 0.0634, 0.0623,
0.0622, 0.063, 0.0633, 0.0628)), .Names = c("x1", "x2", "x3",
"x4", "x5"), class = "data.frame", row.names = c(NA, -25L))
代码如下:
range_span <- function(x) return(diff(range(x))) # function to calculate range
# q6 <- read.table(file="/Users/.../blah.csv",header=T,sep=",") #data
medians <- apply(q6,1,median)
ranges <- apply(q6,1,range_span)
centre <- mean(medians) #grand median
Rtilde <- median(ranges) #median of ranges
plot(medians, type="b",xaxp=c(1, 25, 24),pch=19,xlab="Sample No.",ylab="Medians",main="Median Chart for Thickness of Metal Parts")
# code below draws the control limits
action.limits<-c(centre+0.681*Rtilde,centre-0.681*Rtilde)
warn.limits<-c(centre+(2/3)*0.681*Rtilde,centre-(2/3)*0.681*Rtilde)
abline(h = centre, lty = 3, col = "black")
v0 <-c("CL")
mtext(side = 4, text = v0, at = centre, col = "black", las=2)
abline(h = warn.limits, lty = 3, col = "blue")
v1 <-c("UWL","LWL")
mtext(side = 4, text = v1, at = warn.limits, col = "blue", las=2)
abline(h = action.limits, lty = 3, col = "black")
v2 <-c("UCL","LCL") # the labels for action.limits
mtext(side = 4, text = v2, at = action.limits, col = "black", las=2)
我确信有一个简单的解决方案,我对 RI 完全没有经验,只是想通过在 R 中为课程作业制作图表来给自己设定一个挑战,但现在我的时间已经不多了。
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points()可以帮忙吗?我需要 R 来识别 q6 中的每一行都是一个样本,我可以这样做points(q6,c(1:25))或类似的东西?
【问题讨论】:
-
请使用
dput()发布您的最小可重现示例,请参阅:How to create a Minimal, Complete, and Verifiable example。 -
@EricFail 啊,我现在明白了。感谢您的编辑。
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不客气。我仍然不清楚您想要的行为是什么。您能否以某种方式说明或解释您所看到的情节的确切外观?
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@Eric 完成。有什么想法吗?