Defining group with multiple index value in R

Question

I want to do the Bartlett test with the following data

Sample	RdRp	E	N	Day	Temp
1	103307.0863	118355.7563	288771.7766	3	4
2	107374.2944	119575.9188	266808.8528	3	4
3	102086.9239	119169.198	278197.0355	3	4
4	103994.1128	115458.0308	251387.3436	3	-20
5	103584.6872	114639.1795	253843.8974	3	-20
6	106450.6667	108497.7949	255072.1744	3	-20
7	105704.8475	120569.5917	251049.0129	3	-80
8	107769.3953	117679.2248	249397.3747	3	-80
9	104466.1189	123459.9587	253113.5607	3	-80
10	105046.6462	104636.3077	260975.2615	5	4
11	105867.3231	109970.7077	266720	5	4
12	101763.9385	107098.3385	263026.9538	5	4
13	102250.4304	109121.0127	246532.6582	5	-20
14	103462.8861	123670.481	238045.4684	5	-20
15	96996.4557	113162.5316	254615.6962	5	-20
16	104711.3684	113938.807	242320.5614	5	-80
17	104310.1754	108723.2982	251949.193	5	-80
18	97891.08772	118351.9298	249943.2281	5	-80
19	101149.3671	100744.7696	253682.6127	8	4
20	97507.98987	102363.1595	267034.3291	8	4
21	101553.9646	102767.757	258537.7823	8	4
22	97977.74359	113555.7949	245969.2308	8	-20
23	102897.1282	113145.8462	255808	8	-20
24	109046.359	115195.5897	256627.8974	8	-20
25	105661.883	110944.9771	247899.0331	8	-80
26	106474.6667	111757.7608	250743.7761	8	-80
27	105661.883	116228.0712	258871.6132	8	-80
28	101923.4694	112931.2041	270301.0408	1	25
29	103554.2449	110077.3469	259293.3061	1	25
30	104369.6327	113338.898	242577.8571	1	25
31	58826.07634	57158.11705	130995.827	3	25
32	61185.6285	56425.84224	168016.3868	3	25
33	60860.17303	53496.743	172084.5802	3	25

And I made the following working code.

stab=read.csv("519stability.csv")

bartlett.test(data=stab, RdRp ~ Day, subset = Temp == -80)
bartlett.test(data=stab, RdRp ~ Day, subset = Temp == -20)
bartlett.test(data=stab, RdRp ~ Day, subset = Temp == 4)
bartlett.test(data=stab, RdRp ~ Day, subset = Temp == 25)

bartlett.test(data=stab, E ~ Day, subset = Temp == -80)
bartlett.test(data=stab, E ~ Day, subset = Temp == -20)
bartlett.test(data=stab, E ~ Day, subset = Temp == 4)
bartlett.test(data=stab, E ~ Day, subset = Temp == 25)

bartlett.test(data=stab, N ~ Day, subset = Temp == -80)
bartlett.test(data=stab, N ~ Day, subset = Temp == -20)
bartlett.test(data=stab, N ~ Day, subset = Temp == 4)
bartlett.test(data=stab, N ~ Day, subset = Temp == 25)

This code produces what I want. But is there smart and simple code with some library (dplyr, reshape2, etc.)?

The point is defining a group with two index values; Day and Temp.

---

Edit I add my data with dput() here.

structure(list(Sample = 1:33, RdRp = c(103307.0863, 107374.2944, 
102086.9239, 103994.1128, 103584.6872, 106450.6667, 105704.8475, 
107769.3953, 104466.1189, 105046.6462, 105867.3231, 101763.9385, 
102250.4304, 103462.8861, 96996.4557, 104711.3684, 104310.1754, 
97891.08772, 101149.3671, 97507.98987, 101553.9646, 97977.74359, 
102897.1282, 109046.359, 105661.883, 106474.6667, 105661.883, 
101923.4694, 103554.2449, 104369.6327, 58826.07634, 61185.6285, 
60860.17303), E = c(118355.7563, 119575.9188, 119169.198, 115458.0308, 
114639.1795, 108497.7949, 120569.5917, 117679.2248, 123459.9587, 
104636.3077, 109970.7077, 107098.3385, 109121.0127, 123670.481, 
113162.5316, 113938.807, 108723.2982, 118351.9298, 100744.7696, 
102363.1595, 102767.757, 113555.7949, 113145.8462, 115195.5897, 
110944.9771, 111757.7608, 116228.0712, 112931.2041, 110077.3469, 
113338.898, 57158.11705, 56425.84224, 53496.743), N = c(288771.7766, 
266808.8528, 278197.0355, 251387.3436, 253843.8974, 255072.1744, 
251049.0129, 249397.3747, 253113.5607, 260975.2615, 266720, 263026.9538, 
246532.6582, 238045.4684, 254615.6962, 242320.5614, 251949.193, 
249943.2281, 253682.6127, 267034.3291, 258537.7823, 245969.2308, 
255808, 256627.8974, 247899.0331, 250743.7761, 258871.6132, 270301.0408, 
259293.3061, 242577.8571, 130995.827, 168016.3868, 172084.5802
), Day = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 5L, 5L, 5L, 5L, 
5L, 5L, 5L, 5L, 5L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 8L, 1L, 1L, 
1L, 3L, 3L, 3L), Temp = c(4L, 4L, 4L, -20L, -20L, -20L, -80L, 
-80L, -80L, 4L, 4L, 4L, -20L, -20L, -20L, -80L, -80L, -80L, 4L, 
4L, 4L, -20L, -20L, -20L, -80L, -80L, -80L, 25L, 25L, 25L, 25L, 
25L, 25L)), class = "data.frame", row.names = c(NA, -33L))

Answer 1

You could use an outer approach, where you first put dependent variables and temperatures in vectors and define a bartlett skeleton in a FUNction. Gives a matrix which can easily be splitted into a list. To know which one is which we should setNames.

dpv <- c('RdRp', 'E', 'N')
tmp <- c(-80, -20, 4,  25)
FUN <- Vectorize(\(x, y) 
                 bartlett.test(data=stab, as.formula(paste(x, '~ Day')), 
                               subset=Temp == y), 
                 SIMPLIFY=F)
    
res <- outer(dpv, tmp, FUN) |>
  split(rep(seq(tmp), each=length(dpv))) |>
  setNames(make.names(tmp))

Result

res$X.80  ## the first list element, temp -80
# [[1]]
# 
# Bartlett test of homogeneity of variances
# 
# data:  RdRp by Day
# Bartlett's K-squared = 5.1079, df = 2, p-value = 0.07777
# 
# 
# [[2]]
# 
#   Bartlett test of homogeneity of variances
# 
# data:  E by Day
# Bartlett's K-squared = 0.63384, df = 2, p-value = 0.7284
# 
# 
# [[3]]
# 
# Bartlett test of homogeneity of variances
# 
# data:  N by Day
# Bartlett's K-squared = 1.797, df = 2, p-value = 0.4072

Note: R > 4.1.* is needed.

Answer 2

You can use loops:

library(dplyr)
library(purrr)

map(unique(stab$Temp), ~ bartlett.test(data=stab, RdRp ~ Day, subset = Temp==.x))
map(unique(stab$Temp), ~ bartlett.test(data=stab, E ~ Day, subset = Temp ==.x))
map(unique(stab$Temp), ~ bartlett.test(data=stab, N ~ Day, subset = Temp ==.x))

This may be simplified more with nested loops:

library(purrr)

formulas<-c('RdRp ~ Day', 'E ~ Day', 'N ~ Day')

map(formulas, function(form)
        map(unique(stab$Temp), ~ bartlett.test(data=stab,
                                               as.formula(form),
                                               subset = Temp == .x)))

I can not test this without a sample of your data as code, though.

Answer 3

Yau may try lapply

coln <- c("RdRp", "E", "N")
temps <- c(-80, -20, 4, 25)
lapply(coln, function(x) {
  lapply(temps, function(y){
    bartlett.test(stab[[x]] ~ stab$Day, subset = stab$Temp == y)
  })
  
})

Some part of results are like

[[1]]
[[1]][[1]]

    Bartlett test of homogeneity of variances

data:  stab[[x]] by stab$Day
Bartlett's K-squared = 5.1079, df = 2, p-value = 0.07777


[[1]][[2]]

    Bartlett test of homogeneity of variances

data:  stab[[x]] by stab$Day
Bartlett's K-squared = 2.2096, df = 2, p-value = 0.3313