我有一个函数（weisurv）有2个参数 - sc和shp。 它是通过时间（t）的函数。 时间是一个序列，即t <-seq（1：100）。
weisurv<-function(t,sc,shp){ surv<-exp(-(t/sc)^shp) return(surv) }我有一个数据框（df），其中包含sc和shp值的列表（如300+）。 例如，我有：
M shp sc p C i 1 1 1.138131 10.592154 0.1 1 1 2 1.01 1.143798 10.313217 0.1 1 2 3 1.02 1.160653 10.207863 0.1 1 3 4 1.03 1.185886 9.861997 0.1 1 4 ...我想将sc和shp参数的每一组（ROW）应用到我的函数中。 因此，对于每一行[i]，函数看起来像weisurv（t，sc [[i]]，shp [i]]。 我不明白如何使用apply或adply来做到这一点虽然我确定其中一个或两者的组合都是需要的。 最后，我正在寻找一个数据框，给出一组sc和shp（保持不变的时间），每次给出weisurv的值。 因此，如果我有10组sc和shp参数，我最终会得到10个weisurv时间序列。 谢谢....
I have a function (weisurv) that has 2 parameters - sc and shp. It is a function through time (t). Time is a sequence, i.e. t<-seq(1:100).
weisurv<-function(t,sc,shp){ surv<-exp(-(t/sc)^shp) return(surv) }I have a data frame (df) that contains a list of sc and shp values (like 300+ of them). For example, I have:
M shp sc p C i 1 1 1.138131 10.592154 0.1 1 1 2 1.01 1.143798 10.313217 0.1 1 2 3 1.02 1.160653 10.207863 0.1 1 3 4 1.03 1.185886 9.861997 0.1 1 4 ...I want to apply each set (ROW) of sc and shp parameters to my function. So the function would look like weisurv(t,sc[[i]],shp[i]]) for each row[i]. I do not understand how to use apply or adply to do this though I'm sure one of these or a combo of both are what is needed. In the end, I am looking for a data frame that gives a value of weisurv for each time given a set of sc and shp (held constant through time). So if I had 10 sets of sc and shp parameters, I would end up with 10 time series of weisurv. Thanks....
最满意答案
使用plyr：
作为矩阵（cols中的时间，对应于df行的行）：
aaply(df, 1, function(x) weisurv(t, x$sc, x$shp), .expand = FALSE)作为清单：
alply(df, 1, function(x) weisurv(t, x$sc, x$shp))作为数据框架（按照上面的矩阵结构）：
adply(df, 1, function(x) setNames(weisurv(t, x$sc, x$shp), t))作为长数据帧（每t / sc / shp组合一行）; 注意使用mutate和dplyr的管道运算符）：
newDf <- data.frame(t = rep(t, nrow(df)), sc = df$sc, shp = df$shp) %>% mutate(surv = weisurv(t, sc, shp))您还可以创建一个宽的data.frame，然后使用reshape2::melt重新格式化：
wideDf <- adply(df, 1, function(x) setNames(weisurv(t, x$sc, x$shp), t)) newDf <- melt(wideDf, id.vars = colnames(df), variable.name = "t", value.name = "surv") newDf$t <- as.numeric(as.character(newDf$t))上一个newDf的漂亮情节（使用ggplot2）：
ggplot(newDf, aes(x = t, y = surv, col = sprintf("sc = %0.3f, shp = %0.3f", sc, shp))) + geom_line() + scale_color_discrete(name = "Parameters")
Using plyr:
As a matrix (time in cols, rows corresponding to rows of df):
aaply(df, 1, function(x) weisurv(t, x$sc, x$shp), .expand = FALSE)As a list:
alply(df, 1, function(x) weisurv(t, x$sc, x$shp))As a data frame (structure as per matrix above):
adply(df, 1, function(x) setNames(weisurv(t, x$sc, x$shp), t))As a long data frame (one row per t/sc/shp combination); note uses mutate and the pipe operator from dplyr):
newDf <- data.frame(t = rep(t, nrow(df)), sc = df$sc, shp = df$shp) %>% mutate(surv = weisurv(t, sc, shp))You can also create a wide data.frame and then use reshape2::melt to reformat as long:
wideDf <- adply(df, 1, function(x) setNames(weisurv(t, x$sc, x$shp), t)) newDf <- melt(wideDf, id.vars = colnames(df), variable.name = "t", value.name = "surv") newDf$t <- as.numeric(as.character(newDf$t))Pretty plot of last newDf (using ggplot2):
ggplot(newDf, aes(x = t, y = surv, col = sprintf("sc = %0.3f, shp = %0.3f", sc, shp))) + geom_line() + scale_color_discrete(name = "Parameters")
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