使用条件对所有数字变量进行突变 [英] Mutate all numeric variables with condition
本文介绍了使用条件对所有数字变量进行突变的处理方法，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
我有以下数据集.
structure(list(Monate = structure(c(5L, 4L, 9L, 1L, 8L, 7L, 6L,
2L, 12L, 11L, 10L, 3L), .Label = c("April", "August", "Dezember",
"Februar", "Januar", "Juli", "Juni", "Mai", "März", "November",
"Oktober", "September"), class = "factor"), `06:00:00-00:59:59` = c(1.23650890285152,
1, 0.941959539059369, 0.89618038635695, 0.839845976370379, 0.75789368475892,
0.745357979199247, 0.72726835342115, 0.852384444043199, 0.909567524854388,
0.927385464132519, 0.86987613403725), `06:00:00-08:59:59` = c(0.28594460357425,
0.28900629033325, 0.315719841587, 0.2955345904455, 0.273041491484,
0.26127195229025, 0.23431503049, 0.22167349349575, 0.2603577651915,
0.27084189072775, 0.2653932948955, 0.246001500902), `09:00:00-12:59:59` = c(0.48396300341875,
0.4767075864935, 0.46442852744125, 0.4497871047545, 0.4088477917855,
0.39907331646225, 0.387052062652, 0.3717692228805, 0.39248244105925,
0.422900151297, 0.4111808332435, 0.41687239709375), `13:00:00-16:59:59` = c(0.80304931519625,
0.7588016605465, 0.66548492801425, 0.6354014724685, 0.670969199053,
0.58076568424625, 0.67418744877975, 0.5756792489565, 0.66766765226025,
0.642654220031, 0.665735145053, 0.64571729495575), `17:00:00-19:59:59` = c(1.44101730420325,
1.35385679874675, 1.2576760885645, 1.1478274405785, 1.04871946565175,
0.96498516399125, 0.9775891903885, 0.9959819200865, 1.0805495717725,
1.2042836233065, 1.29337213775425, 1.17905118172475), `20:00:00-22:59:59` = c(2.7590124723295,
2.1458514208575, 2.00826846019375, 1.91719758837025, 1.7425652515095,
1.509920737758, 1.3851778869395, 1.43588733645025, 1.8753361739725,
2.03981388252275, 2.013661241837, 1.79544669375925), `23:00:00-00:59:59` = c(1.92548355216275,
0.926693335582, 0.921407232611, 0.92680356666725, 0.87033626954225,
0.81832375651025, 0.75033232500425, 0.7289554638715, 0.79565322269175,
0.85606802035675, 0.90891474021375, 0.94322847859325)), class = "data.frame", row.names = c(NA,
-12L))
我想建立将一个月设置为"1"的索引值.
I want to build index values where I set one month to "1".
saisonindex <- RWavg %>%
mutate(`06:00:00-00:59:59` = `06:00:00-00:59:59` / `06:00:00-00:59:59`[Monate == "Februar"])
saisonindex
Monate 06:00:00-00:59:59 06:00:00-08:59:59 09:00:00-12:59:59 13:00:00-16:59:59 17:00:00-19:59:59 20:00:00-22:59:59 23:00:00-00:59:59
1 Januar 1.2365089 0.2859446 0.4839630 0.8030493 1.4410173 2.759012 1.9254836
2 Februar 1.0000000 0.2890063 0.4767076 0.7588017 1.3538568 2.145851 0.9266933
3 März 0.9419595 0.3157198 0.4644285 0.6654849 1.2576761 2.008268 0.9214072
4 April 0.8961804 0.2955346 0.4497871 0.6354015 1.1478274 1.917198 0.9268036
5 Mai 0.8398460 0.2730415 0.4088478 0.6709692 1.0487195 1.742565 0.8703363
6 Juni 0.7578937 0.2612720 0.3990733 0.5807657 0.9649852 1.509921 0.8183238
7 Juli 0.7453580 0.2343150 0.3870521 0.6741874 0.9775892 1.385178 0.7503323
8 August 0.7272684 0.2216735 0.3717692 0.5756792 0.9959819 1.435887 0.7289555
9 September 0.8523844 0.2603578 0.3924824 0.6676677 1.0805496 1.875336 0.7956532
10 Oktober 0.9095675 0.2708419 0.4229002 0.6426542 1.2042836 2.039814 0.8560680
11 November 0.9273855 0.2653933 0.4111808 0.6657351 1.2933721 2.013661 0.9089147
12 Dezember 0.8698761 0.2460015 0.4168724 0.6457173 1.1790512 1.795447 0.9432285
我想对所有数字列动态执行此操作，我想mutate_if(is.numeric)会执行此操作，但是我无法理解括号中的条件可能如何工作.
I want to do this for all numeric columns dynamically, I guess mutate_if(is.numeric) will do this, but I can not get my head around how it might work which the condition in the bracket.
你能帮我吗?
推荐答案
使用mutate_if，指定~，然后对"Monate"为"Februar"的列值进行子集
Using mutate_if, specify the ~ and then subset the column value where the 'Monate' is "Februar"
library(dplyr)
RWavg %>%
mutate_if(is.numeric, ~ ./.[Monate == "Februar"])
在base R中，仅对行进行子集处理，正确确定长度，划分并更新列
In base R, just subset the row, get the lengths right, divide and update the columns
RWavg[-1] <- RWavg[-1]/RWavg[RWavg$Monate == 'Februar', -1][col(RWavg[-1])]
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