R在数据框中创建希腊符号

Question

我正在尝试创建一个包含希腊符号的自定义方面标题列，用于 ggplot。但是，由于某种原因，函数 expression() 没有像我预期的那样工作。

如果我用

df$fac <- ifelse(df$hour==1,expression(paste(df$hour+11, " - ", df$hour, " am; ", mu, 't=', round(df$exp), sep="")), "k")

结果是

> tail(df,7)
   hour     exp duration                                                               fac
18    1 3721141 921184.3            expression(paste(df$hour + 11, " - ", df$hour, " am; ", 
19    2 2175734 921844.5            mu, "t=", round(df$exp), sep = ""), "k", "k", "k", "k", 
20   24 6656036 919250.4                                                          "k", "k")
21    6 7869344 918690.6                                                               <NA>
22    4 2220189 920862.5                                                               <NA>
23    5 2780161 920005.2                                                               <NA>
24    3 2690003 922332.2                                                               <NA>
Warning message:
In format.data.frame(x, digits = digits, na.encode = FALSE) :
  corrupt data frame: columns will be truncated or padded with NAs

如果我使用

df$fac <- ifelse(df$hour==1,paste(df$hour+11, " - ", df$hour, " am; ",expression(mu), 't=', round(df$exp), sep=""), "k")

它给了我

> tail(df,7)
   hour     exp duration                    fac
18    1 3721141 921184.3 12 - 1 am; mut=3721141
19    2 2175734 921844.5                      k
20   24 6656036 919250.4                      k
21    6 7869344 918690.6                      k
22    4 2220189 920862.5                      k
23    5 2780161 920005.2                      k
24    3 2690003 922332.2                      k

如您所见，这两种情况都没有在数据框中给我一个希腊符号。我也用了bquote()也没有用。名为 fac 的列实际上应该存储构面标签。例如，对于第 1 小时，格式应为 12 - 1 am; μt= 3721141 你能告诉我正确的方法吗？

这是示例数据

> dput(df)
structure(list(hour = c(14L, 16L, 17L, 19L, 21L, 11L, 18L, 22L, 
8L, 12L, 23L, 9L, 15L, 13L, 7L, 10L, 20L, 1L, 2L, 24L, 6L, 4L, 
5L, 3L), exp = c(3923432.39149989, 5109851.42312691, 7070493.23531211, 
20803629.8380627, 11256055.2295972, 4757385.54314779, 27576092.3667234, 
9141934.32914581, 35348337.463401, 4108302.35295023, 6489550.8228499, 
9848278.71615835, 3816330.82689788, 3790223.01364096, 26886283.3156086, 
6308698.90957706, 17434998.9294133, 3721141.03502774, 2175733.94348253, 
6656035.63316213, 7869343.75835358, 2220189.45505708, 2780160.84185653, 
2690003.082563), duration = c(861081.7025, 865706.1025, 872852.945833333, 
892462.086388889, 908292.911111111, 862927.383055556, 882474.983055556, 
913184.621388889, 895447.985277778, 861349.923888889, 916578.3325, 
878670.013611111, 862413.049722222, 861035.062777778, 911718.763055556, 
868130.278333333, 901928.8125, 921184.328055556, 921844.501111111, 
919250.422777778, 918690.569722222, 920862.491111111, 920005.158333333, 
922332.197777778)), .Names = c("hour", "exp", "duration"), row.names = c(NA, 
-24L), class = "data.frame")

Answer 1

不要尝试在 data.frame 中放入表达式 - 您可以包含要解析的文本向量，然后使用 labeller=label_parsed:

library(ggplot2)
d <- data.frame(x=runif(20), y=runif(20), grp=rep(c('mu', 'sigma'), 10))
ggplot(d, aes(x=x, y=y)) +
  geom_point() +
  facet_wrap(~grp, labeller=label_parsed)

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