如何以长格式的R数据帧的子集进行操作？

本文介绍了如何以长格式的R数据帧的子集进行操作？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 我有一个3组和3天的数据框架： set.seed（10） dat < - data.frame（group = rep（c（g1，g2，g3），each = 3），day = rep（c（0,2,4），3），value = （9））＃组日值＃1 g1 0 0.507478 ＃2 g1 2 0.306769 ＃3 g1 4 0.426908 ＃4 g2 0 0.693102 ＃5 g2 2 0.085136 ＃6 g2 4 0.225437 ＃7 g3 0 0.274531 ＃8 g3 2 0.272305 ＃9 g3 4 0.615829 我想采用log2并将每个值与每个组中的第0个值进行分隔。我现在的做法是通过在中间步骤中计算每个日组： day_0< - dat [dat $ day == 0，value] day_2< - dat [dat $ day == 2，value] day_4< - dat [dat $ day == 4，value ] res rownames（res）< - c（g1，g2 g3） colnames（res）< - c（day_0，log_ratio_day_2_day_0，log_ratio_day_4_day_0）＃day_0 log_ratio_day_2_day_0 log_ratio_day_4_day_0 ＃g1 0 -0.7261955 -0.249422 ＃g2 0 -3.0252272 -1.620346 ＃g3 0 -0.0117427 1.165564 什么是正确的解决方案你的朋友是在...之间计算 res 包含 c code code code code code $ c $ require（plyr）> ddply（dat，。（group），mutate，new_value = log2（value / value [1]））组日值new_value 1 g1 0 0.50747820 0.00000000 2 g1 2 0.30676851 -0.72619548 3 g1 4 0.42690767 -0.24942179 4 g2 0 0.69310208 0.00000000 5 g2 2 0.08513597 -3.02522716 6 g2 4 0.22543662 -1.62034599 7 g3 0 0.27453052 0.00000000 8 g3 2 0.27230507 -0.01174274 9 g3 4 0.61582931 1.16556397 I have a data frame with 3 groups and 3 days:set.seed(10)dat <- data.frame(group=rep(c("g1","g2","g3"),each=3), day=rep(c(0,2,4),3), value=runif(9))# group day value# 1 g1 0 0.507478# 2 g1 2 0.306769# 3 g1 4 0.426908# 4 g2 0 0.693102# 5 g2 2 0.085136# 6 g2 4 0.225437# 7 g3 0 0.274531# 8 g3 2 0.272305# 9 g3 4 0.615829I want to take the log2 and divide each value with the day 0 value within each group. The way I'm doing it now is by calculating each day group in an intermediate step:day_0 <- dat[dat$day==0, "value"]day_2 <- dat[dat$day==2, "value"]day_4 <- dat[dat$day==4, "value"]res <- cbind(0, log2(day_2/day_0), log2(day_4/day_0))rownames(res) <- c("g1","g2","g3")colnames(res) <- c("day_0","log_ratio_day_2_day_0","log_ratio_day_4_day_0")# day_0 log_ratio_day_2_day_0 log_ratio_day_4_day_0# g1 0 -0.7261955 -0.249422# g2 0 -3.0252272 -1.620346# g3 0 -0.0117427 1.165564What's the proper way of calculating res without an intermediate step? 解决方案 Your friend is ddply from the plyr package:require(plyr)> ddply(dat, .(group), mutate, new_value = log2(value / value[1])) group day value new_value1 g1 0 0.50747820 0.000000002 g1 2 0.30676851 -0.726195483 g1 4 0.42690767 -0.249421794 g2 0 0.69310208 0.000000005 g2 2 0.08513597 -3.025227166 g2 4 0.22543662 -1.620345997 g3 0 0.27453052 0.000000008 g3 2 0.27230507 -0.011742749 g3 4 0.61582931 1.16556397 这篇关于如何以长格式的R数据帧的子集进行操作？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！