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I would like to add another level of complexity to the color coding scheme I have going on in the below plot. I want to account for whether each of the values being plotted has passed a statistical test. So, the dots will only be color coded based on the percentile if they pass the test, otherwise, I would like the dot to be grey.

Here is my code as I have it after all the helpful suggestions I received from my first post Color code points based on percentile in ggplot (note: this is some made up data, though I have real data which has many more entries:

dat <- data.frame(key = c("a1-a3", "a1-a2"), position = 1:100, fst = rlnorm(200, 0, 1), fet = rnorm(200, 0.24, 0.54))

#Get quantiles
quants <- quantile(dat$fst, c(0.95, 0.99))

dat$quant  <- with(dat, factor(ifelse(fst < quants[1], 0,
                                  ifelse(fst < quants[2], 1, 2))))

dat$fisher <- with(dat, factor(ifelse(fet > 1.30102999566398, 0, 1)))

dat$col <- with(dat, factor(ifelse(fet < 1.30102999566398, 3, quant)))

########theme set
theme_set(theme_bw(base_size = 10))

p1 <- ggplot(dat, aes(x=position, y=fst)) +
  geom_point(aes(colour = col, size=0.2)) +
  facet_wrap(~key, nrow = 1) +
  scale_colour_manual(values = c("black", "blue", "red", "grey"), labels = c("0-95", "95-99", "99-100", "fail")) +
  ylab(expression(F[ST])) +
  xlab("Genomic Position (Mb)") +
  scale_x_continuous(breaks=c(0, 1e+06, 2e+06, 3e+06, 4e+06), labels=c("0", "1", "2", "3", "4")) +
  scale_y_continuous(limits=c(0,1)) +
  theme(plot.background = element_blank(),
    panel.background = element_blank(),
    panel.border = element_blank(),
    legend.position="none",
    legend.title = element_blank()
    )

tiff(Fstvalues_colourcode3.tiff", height=2.5, width=6.5, units="in", res = 300, pointsize="10")
p1
dev.off()

My problem is in the line: dat$col <- with(dat, factor(ifelse(fet < 1.30102999566398, 3, quant))). I want it to use the value from the $quant if it has an $fet value above the above listed value (or fisher == 0), and if it has an $fet value below, I would like it to make a new factor (3). When I look at the data frame it is doing something different than this. Any comments/suggestions are much appreciated! (I'm pretty new to coding and am finding factors are not easy to work with!!)

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1 回答 1

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是的，你是对的，with(dat, factor(ifelse(fet < 1.30102999566398, 3, quant)))给出了一个“意外”的结果。您在quant 中的no返回值被强制转换为与返回值 (3) a相同的类。看看：ifelsefactoryesnumerictail(dat[order(dat$fet), c("fet", "quant", "col")])

#          fet quant col
# 6   1.202582     0   3
# 40  1.318997     0   1
# 74  1.324552     0   1
# 24  1.415189     1   2
# 38  1.418230     0   1
# 123 1.531584     0   1

对于 fet > 1.301 ( testin ifelse)，'col' 变为 1, 1, 2, 1, 1，而不是 0, 0, 1, 0, 0。这样的事情发生了： 

# original factor version of quant
quant <- as.factor(0:2)
quant
# [1] 0 1 2
# Levels: 0 1 2

# coerce quant to numeric
as.numeric(quant)
# [1] 1 2 3

比较这两个：

set.seed(1)
df <- data.frame(fet = rnorm(9), quant = factor(0:2))
str(df)
df$col <- with(df, ifelse(fet < 0, 3, quant))
df

set.seed(1)
df <- data.frame(fet = rnorm(9), quant = 0:2)
str(df)
df$col <- with(df, ifelse(fet < 0, 3, quant))
df

因此，请尝试factor从您ifelse创建“量化”的呼叫中删除，看看它是否解决了问题。

另请参阅此处的 8.2.1：http ://www.burns-stat.com/pages/Tutor/R_inferno.pdf 。

PS。当您提出问题时，单行ifelse是您的实际问题（不是绘图部分）。如果是这样，您可能希望隔离此问题并浓缩您的问题。

于 2013-10-12T12:44:30.673 回答