Since last version, mlr3tuning package supports (custom) instantiated resampling in AutoTuner class: https://github.com/mlr-org/mlr3tuning/releases/tag/v0.17.2
I have tried to construct rolling window CV's with custom resmapling as in the following link https://towardsdatascience.com/time-series-nested-cross-validation-76adba623eb9 (figure 1).
I want to tune hyperparameters on (let's say )row ids 1:1000 (for example 1:800 train and 801:100 test). and then I would like to evaluate the model on test set, say 1001:1100.
Her is my try:
library(mlr3)
library(mlr3tuning)
library(mlr3pipelines)
library(mlr3learners)
# task
task = tsk("iris")
task_ = task$clone()
data_ = task_$data()
data_ = cbind(data_, monthid = c(rep(1, 30), rep(2, 30), rep(3, 30), rep(4, 30), rep(5, 30)))
task = as_task_classif(data_, target = "Species")
# inner custom rolling window resampling
custom = rsmp("custom")
task_ <- task$clone()
task_$set_col_roles("monthid", "group")
groups = task_$groups
rm(task_)
groups_v <- groups[, unique(group)]
train_length <- 2
test_length <- 1
train_groups <- lapply(0:(length(groups_v)-(train_length+1)), function(x) x + (1:train_length))
test_groups <- lapply(train_groups, function(x) tail(x, 1) + test_length)
train_sets <- lapply(train_groups, function(x) groups[group %in% groups_v[x], row_id])
test_sets <- lapply(test_groups, function(x) groups[group %in% groups_v[x], row_id])
custom$instantiate(task, train_sets, test_sets)
# outer custom rolling window resampling
customo = rsmp("custom")
task_ <- task$clone()
task_$set_col_roles("monthid", "group")
groups = task_$groups
rm(task_)
groups_v <- groups[, unique(group)]
train_length_out <- train_length + test_length
test_length_out <- 1
train_groups_out <- lapply(0:(length(groups_v)-(train_length_out+1)), function(x) x + (1:train_length_out))
test_groups_out <- lapply(train_groups_out, function(x) tail(x, 1) + test_length_out)
train_sets_out <- lapply(train_groups_out, function(x) groups[group %in% groups_v[x], row_id])
test_sets_out <- lapply(test_groups_out, function(x) groups[group %in% groups_v[x], row_id])
customo$instantiate(task, train_sets_out, test_sets_out)
# inspect custom cv's
custom$train_set(1)
custom$test_set(1)
(max(custom$train_set(1)) + 1) == head(custom$test_set(1), 1) # test set starts after train set
customo$train_set(1)
customo$test_set(1)
(max(customo$train_set(1)) + 1) == head(customo$test_set(1), 1) # test set starts after train set
all(c(custom$train_set(1), custom$test_set(1)) %in% customo$train_set(1)) # first outer set contains all inner sets
length(intersect(customo$test_set(1), c(custom$train_set(1), custom$test_set(1)))) == 0
# costruct graph
graph = po("removeconstants", id = "removeconstants_1", ratio = 0) %>>%
po("branch", options = c("nop_prep", "yeojohnson", "pca", "ica"), id = "prep_branch") %>>%
gunion(list(po("nop", id = "nop_prep"), po("yeojohnson"), po("pca", scale. = TRUE), po("ica"))) %>>%
po("unbranch", id = "prep_unbranch") %>>%
po("learner", learner = lrn("classif.rpart"))
plot(graph)
graph_learner = as_learner(graph)
as.data.table(graph_learner$param_set)[1:70, .(id, class, lower, upper)]
search_space = ps(
prep_branch.selection = p_fct(levels = c("nop_prep", "yeojohnson", "pca", "ica")),
pca.rank. = p_int(2, 6, depends = prep_branch.selection == "pca"),
ica.n.comp = p_int(2, 6, depends = prep_branch.selection == "ica"),
yeojohnson.standardize = p_lgl(depends = prep_branch.selection == "yeojohnson")
)
at = auto_tuner(
method = "random_search",
learner = graph_learner,
resampling = custom,
measure = msr("classif.acc"),
search_space = search_space
)
# resmpling
rr = resample(task, at, customo, store_models = TRUE)
I get an error:
INFO [09:46:49.340] [mlr3] Applying learner 'removeconstants_1.prep_branch.nop_prep.yeojohnson.pca.ica.prep_unbranch.classif.rpart.tuned' on task 'data_' (iter 1/2)
INFO [09:46:49.362] [mlr3] Applying learner 'removeconstants_1.prep_branch.nop_prep.yeojohnson.pca.ica.prep_unbranch.classif.rpart.tuned' on task 'data_' (iter 2/2)
Error: Train set 3 of inner resampling 'custom' contains row ids not present in task 'data_': {91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120}
The first outer training set contains these ids:
train_sets_out
#> [[1]]
#> [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90
This means row 1 to 90 are available in the inner resampling.
The third training set of the inner resampling needs rows 61 to 120 but rows 91 to 120 are not available.
train_sets
#> [[3]]
#> [1] 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120
So that can't work. Check out figure 4.7 in the book to see how the outer and inner resampling work together.
Related
I have some dataframe. Here is a small expample:
a <- rnorm(100, 5, 2)
b <- rnorm(100, 10, 3)
c <- rnorm(100, 15, 4)
df <- data.frame(a, b, c)
And I have a character variable vect <- "c('a','b')"
When I try to calculate sum of vars using command
df$d <- df[vect]
which must be an equivalent of
df$d <- df[c('a','b')]
But, as a reslut I have got an error
[.data.frame(df, vect) :undefined columns selected
You're assumption that
vect <- "c('a','b')"
df$d <- df[vect]
is equivalent to
df$d <- df[c('a','b')]
is incorrect.
As #Karthik points out, you should remove the quotation marks in the assignment to vect
However, from your question it sounds like you want to then sum the elements specified in vect and then assign to d. To do this you need to slightly change your code
vect <- c('a','b')
df$d <- apply(X = df[vect], MARGIN = 1, FUN = sum)
This does elementwise sum on the columns in df specified by vect. The MARGIN = 1 specifies that we want to apply the sum rowise rather than columnwise.
EDIT:
As #ThomasIsCoding points out below, if for some reason vect has to be a string, you can parse a string to an R expression using str2lang
vect <- "c('a','b')"
parsed_vect <- eval(str2lang(vect))
df$d <- apply(X = df[parsed_vect], MARGIN = 1, FUN = sum)
Perhaps you can try
> df[eval(str2lang(vect))]
a b
1 8.1588519 9.0617818
2 3.9361214 13.2752377
3 5.5370983 8.8739725
4 8.4542050 8.5704234
5 3.9044461 13.2642793
6 5.6679639 12.9529061
7 4.0183808 6.4746806
8 3.6415608 11.0308990
9 4.5237453 7.3255129
10 6.9379168 9.4594150
11 5.1557935 11.6776181
12 2.3829337 3.5170335
13 4.3556430 7.9706624
14 7.3274615 8.1852829
15 -0.5650641 2.8109197
16 7.1742283 6.8161200
17 3.3412044 11.6298940
18 2.5388981 10.1289533
19 3.8845686 14.1517643
20 2.4431608 6.8374837
21 4.8731053 12.7258259
22 6.9534912 6.5069513
23 4.4394807 14.5320225
24 2.0427553 12.1786148
25 7.1563978 11.9671603
26 2.4231207 6.1801862
27 6.5830372 0.9814878
28 2.5443326 9.8774632
29 1.1260322 9.4804636
30 4.0078436 12.9909014
31 9.3599808 12.2178596
32 3.5362245 8.6758910
33 4.6462337 8.6647953
34 2.0698037 7.2750532
35 7.0727970 8.9386798
36 4.8465248 8.0565347
37 5.6084462 7.5676308
38 6.7617479 9.5357666
39 5.2138482 13.6822924
40 3.6259103 13.8659939
41 5.8586547 6.5087016
42 4.3490281 9.5367522
43 7.5130701 8.1699117
44 3.7933813 9.3241308
45 4.9466813 9.4432584
46 -0.3730035 6.4695187
47 2.0646458 10.6511916
48 4.6027309 4.9207746
49 5.9919348 7.1946723
50 6.0148330 13.4702419
51 5.5354452 9.0193366
52 5.2621651 12.8856488
53 6.8580210 6.3526151
54 8.0812166 14.4659778
55 3.6039030 5.9857886
56 9.8548553 15.9081336
57 3.3675037 14.7207681
58 3.9935336 14.3186175
59 3.4308085 10.6024579
60 3.9609624 6.6595521
61 4.2358603 10.6600581
62 5.1791856 9.3241118
63 4.6976289 13.2833055
64 5.1868906 7.1323826
65 3.1810915 12.8402472
66 6.0258287 9.3805249
67 5.3768112 6.3805096
68 5.7072092 7.1130150
69 6.5789349 8.0092541
70 5.3175820 17.3377234
71 9.7706112 10.8648956
72 5.2332127 12.3418373
73 4.7626124 13.8816910
74 3.9395911 6.5270785
75 6.4394724 10.6344965
76 2.6803695 10.4501753
77 3.5577834 8.2323369
78 5.8431140 7.7932460
79 2.8596818 8.9581837
80 2.7365174 10.2902512
81 4.7560973 6.4555758
82 4.6519084 8.9786777
83 4.9467471 11.2818536
84 5.6167284 5.2641380
85 9.4700525 2.9904731
86 4.7392906 11.3572521
87 3.1221908 6.3881556
88 5.6949432 7.4518023
89 5.1435241 10.8912283
90 2.1628966 10.5080671
91 3.6380837 15.0594135
92 5.3434709 7.4034042
93 -0.1298439 0.4832707
94 7.8759390 2.7411723
95 2.0898649 9.7687250
96 4.2131549 9.3175228
97 5.0648105 11.3943350
98 7.7225193 11.4180456
99 3.1018895 12.8890257
100 4.4166832 10.4901303
I have data with a grouping variable 'grps' and a value 'x'. I have calculated the cummax within each group 'cmx'. Now I need to find the second highest cumulative value of 'x' within each group, scmx.
Some data, including the desired column scmx:
library(data.table)
d = structure(list(date = structure(rep(c(18690, 18691, 18692, 18693, 18694, 18695, 18696, 18697), 2), class = "Date"),
x = c(18, 70, 57, 94, 94, 13, 98, 23, 20, 72, 59, 96, 96, 15, 100, 25),
grps = c(rep("g1", 8), rep("g2", 8))),
row.names = c(NA, -16L), class = c("data.table", "data.frame"))
d[, cmx := cummax(x), by = .(grps)]
d[, scmx := c(18, 18, 57, 70, 70, 70, 94, 94, 20, 20, 59, 72, 72, 72, 96, 96)]
Context
If x corresponds to a performance rating, what I am trying to do is locate the date when they achieved their best performance and their second best. A similar question of mine where I needed to locate the row which corresponded to the highest cumulative value in a column:
Fill down first row within each cumulative max, with a twist
A data.table alternative:
d[ , scmx2 := {
c(x[1], sapply(seq(.N)[-1], function(i){
v = x[1:i]
v[frank(-v, ties.method = "dense") == 2][1]
}))
}, by = grps]
# date x grps cmx scmx scmx2
# 1: 2021-03-04 18 g1 18 18 18
# 2: 2021-03-05 70 g1 70 18 18
# 3: 2021-03-06 57 g1 70 57 57
# 4: 2021-03-07 94 g1 94 70 70
# 5: 2021-03-08 94 g1 94 70 70
# 6: 2021-03-09 13 g1 94 70 70
# 7: 2021-03-10 98 g1 98 94 94
# 8: 2021-03-11 23 g1 98 94 94
# 9: 2021-03-04 20 g2 20 20 20
# 10: 2021-03-05 72 g2 72 20 20
# 11: 2021-03-06 59 g2 72 59 59
# 12: 2021-03-07 96 g2 96 72 72
# 13: 2021-03-08 96 g2 96 72 72
# 14: 2021-03-09 15 g2 96 72 72
# 15: 2021-03-10 100 g2 100 96 96
# 16: 2021-03-11 25 g2 100 96 96
Within each group (by = grps), loop (sapply) over a sequence from 2 to number of rows in the current group (seq(.N)[-1]). In each step, subset 'x' from start of the vector to the index 'i' (v = x[1:i]).
Calculate dense rank and check if the rank is 2 (frank(-v, ties.method = "dense") == 2), i.e. the rank of the second largest number. Use the logical indices to subset 'v' (v[...). Select the first match ([1]; in case of several values with rank 2). Concatenate the result from this 'expanding window' with the first element of 'x' (c(x[1], ...).
In the first window, with only one value, there is clearly no second highest value. Here OP have chosen to return the first value. The same choice needs to be made also for longer windows where all values are equal, which will occur when there are leading runs of equal values. If we rather want to return NA than the first value, then replace the x[1] in the line
c(x[1], sapply(seq(.N)[-1], function(i){
...with NA_real_.
Small demo:
d = data.table(grps = c(1, 1, 2, 2, 2), x = c(3, 3, 4, 4, 5))
d[ , scmx2 := {
c(NA_real_, sapply(seq(.N)[-1], function(i){
v = x[1:i]
v[frank(-v, ties.method = "dense") == 2][1]
}))
}, by = grps]
# grps x scmx
# 1: 1 3 NA # grp 1: all values equal in all windows -> all NA
# 2: 1 3 NA
# 3: 2 4 NA
# 4: 2 4 NA
# 5: 2 5 4 # grp 2: only the last window has a second highest value
This question is indeed similar to the post I linked to above (Finding cumulative second max per group in R). However, here OP asked for a data.table solution.
Here is another option using non-equi join:
d[, s2 := .SD[.SD, on=.(grps, date<=date, x<cmx), by=.EACHI, max(x.x)]$V1]
d[is.na(s2), s2 := x][]
output:
date x grps cmx scmx s2
1: 2021-03-04 18 g1 18 18 18
2: 2021-03-05 70 g1 70 18 18
3: 2021-03-06 57 g1 70 57 57
4: 2021-03-07 94 g1 94 70 70
5: 2021-03-08 94 g1 94 70 70
6: 2021-03-09 13 g1 94 70 70
7: 2021-03-10 98 g1 98 94 94
8: 2021-03-11 23 g1 98 94 94
9: 2021-03-04 20 g2 20 20 20
10: 2021-03-05 72 g2 72 20 20
11: 2021-03-06 59 g2 72 59 59
12: 2021-03-07 96 g2 96 72 72
13: 2021-03-08 96 g2 96 72 72
14: 2021-03-09 15 g2 96 72 72
15: 2021-03-10 100 g2 100 96 96
16: 2021-03-11 25 g2 100 96 96
Create a sequence that is the length of the column x. Apply the function to each sequence in x that is from index 1 to the current number in the sequence, only caring about the unique values. Rfast::nth can be used to take the 2nd highest number in a vector.
library(Rfast)
sapply(seq(length(d$x)), function(x) {
return(nth(unique(d$x[1:x]), 2, descending=TRUE))
})
[1] 2.652495e-315 1.800000e+01 5.700000e+01 7.000000e+01
[5] 7.000000e+01 7.000000e+01 9.400000e+01 9.400000e+01
To do it for the new data frame. We can still use the function created above. Arrange the data frame so that the group names and values are in their own column, then use lapply with rollapplyr to capture the 2nd largest unique value.
d1=d %>% select(-cmx) %>%
pivot_wider(names_from=grps, values_from=x)
lapply(d1[-1], function(x) {
my_list=rollapplyr(x, seq(length(x)), function(x) {return(nth(sort(unique(x), decreasing=TRUE), 2))})
return(my_list)
})
In R, I have a df such as:
a b c
1 124 70 aa
2 129 67 aa
3 139 71 aa
4 125 77 aa
5 125 82 aa
6 121 69 aa
7 135 68 bb
8 137 72 bb
9 137 78 bb
10 140 86 bb
I want to iterate along rows within columns (a, b), computing the mean of all rows pairs, and paste this mean to the same two rows of new columns (a_new, b_new) if the difference between these two rows is >=12. Otherwise just copy the old value. This behaviour should be restricted to groups as marked by another column (c), i.e it should not happen if two rows are from different groups.
In this example, it happens in row 3 (cos in column a, difference with next (4th) row is 14) and in row 5 (cos in column b, difference with next row is 13). However, this should not happen with row 6 cos row 7 is in another c group.
Thus, resulting df would look like:
a b c a_new b_new
1 124 70 aa 124 70
2 129 67 aa 129 67
3 139 71 aa 132 71
4 125 77 aa 132 68
5 125 82 aa 125 75.5
6 121 69 aa 121 75.5
7 135 68 bb 135 68
8 137 72 bb 137 72
9 137 78 bb 137 78
10 140 86 bb 140 86
I've been struggling to do this for a while, figured out that perhaps lag function could be used, but no success. Help would be much appreciated (be it base R, or dplyr, or whatever)
Dput:
structure(list(a = c(124, 129, 139, 125, 125, 121, 135, 137,
137, 140), b = c(70, 67, 71, 77, 82, 69, 68, 72, 78, 86), c = c("aa",
"aa", "aa", "aa", "aa", "aa", "bb", "bb", "bb", "bb")), row.names = c(NA,
-10L), class = c("tbl_df", "tbl", "data.frame"))
We can write a function which works for one chunk.
apply_fun <- function(x) {
inds <- which(abs(diff(x)) >= 12)
if(length(inds))
x[sort(c(inds, inds + 1))] <- c(sapply(inds, function(i)
rep(mean(x[c(i, i + 1)]), 2)))
return(x)
}
and then apply it for multiple columns by group.
library(dplyr)
df %>% group_by(c) %>% mutate_at(vars(a, b), list(new = apply_fun))
# a b c a_new b_new
# <dbl> <dbl> <chr> <dbl> <dbl>
# 1 124 70 aa 124 70
# 2 129 67 aa 129 67
# 3 139 71 aa 132 71
# 4 125 77 aa 132 77
# 5 125 82 aa 125 75.5
# 6 121 69 aa 121 75.5
# 7 135 68 bb 135 68
# 8 137 72 bb 137 72
# 9 137 78 bb 137 78
#10 140 86 bb 140 86
What I understood is to apply to each group given by the indicator column "c" the procedure commented in the code below:
pairAverage <- function(x) {
# x should be a numeric vector of length > 1
if (is.vector(x) & is.numeric(x) & length(x) > 1) {
# copy data to an aux vector
aux <- x
# get differences of lag 1
dh<-diff(x, 1)
# get means of consecutive pairs
med <- c(x$a[2:length(x)] - dh/2)
# get positions (index) of abs(means) >= 12
idx <- match(med[abs(dh) >= 12], med)
# need 2 reps of each mean to replace consecutive values of x
valToRepl <- med[sort(rep(idx,2))]
# ordered indexes pairs of consecutive elements of x to be replaced
idxToRepl <- sort(c(idx,idx+1))
# replace pairs of values
aux[idxToRepl] <- valToRepl
return(aux)
} else {
# do nothing
warning("paramater x should be a numeric vector of length > 1")
return(NULL)
}
}
pairAverageByGroups <- function(x, gr) {
if (is.vector(x) & is.numeric(x) & length(x) == length(gr)) {
x.ls <- split(x, as.factor(gr))
output <- unlist(lapply(x.ls, pairAverage))
names(output) <- NULL
output
} else {
# do nothing
warning("paremater x should be a numeric vector of length > 1")
return(NULL)
}
}
pairAverageByGroups(dd$a, dd$c)
[1] 124 129 132 132 125 121 135 137 137 140
my dataset is :
x=data.frame(v1=c(97 , 97 , 85 , 84 , 90 , 80 , 81 , 90 , 80, 70, 90 , 90, 90 ,95 , 88 , 99),
+ v2=c(99 , 91 , 91 ,83 , 99 , 95 , 74 , 88 , 82 , 80 , 96 , 87 , 92 , 96 , 88, 95),
+ v3=c( 89 ,93 , 87 , 80 , 96 , 96 , 75 , 90 , 78, 86 , 92 ,88 , 80, 88 , 98 ,98),
+ v4=c( 89 , 97 ,91 , 86 , 95 , 95 , 89 , 88 , 75, 82 , 99, 92 , 95, 92 , 90, 98),
+ v5=c( 99 ,90 , 93 ,91 , 90 , 90 , 77 , 92 , 85, 76 , 90, 96 , 90, 90 , 90, 92))
> x
v1 v2 v3 v4 v5
1 97 99 89 89 99
2 97 91 93 97 90
3 85 91 87 91 93
4 84 83 80 86 91
5 90 99 96 95 90
6 80 95 96 95 90
7 81 74 75 89 77
8 90 88 90 88 92
9 80 82 78 75 85
10 70 80 86 82 76
11 90 96 92 99 90
12 90 87 88 92 96
13 90 92 80 95 90
14 95 96 88 92 90
15 88 88 98 90 90
16 99 95 98 98 92
I used rpart package to apply decision tree as follows :
# Classification Tree with rpart
library(rpart)
fit <- rpart(v5 ~ v1+v2+v3+v4,
method="class", data=x)
printcp(fit) # display the results
Classification tree:
rpart(formula = v5 ~ v1 + v2 + v3 + v4, data = x, method = "class")
Variables actually used in tree construction:
character(0)
Root node error: 9/16 = 0.5625
n= 16
CP nsplit rel error xerror xstd
1 0.01 0 1 0 0
> summary(fit) # detailed summary of splits
Call:
rpart(formula = v5 ~ v1 + v2 + v3 + v4, data = x, method = "class")
n= 16
CP nsplit rel error xerror xstd
1 0.01 0 1 0 0
Node number 1: 16 observations
predicted class=90 expected loss=0.5625 P(node) =1
class counts: 1 1 1 7 1 2 1 1 1
probabilities: 0.062 0.062 0.062 0.438 0.062 0.125 0.062 0.062 0.062
plot tree
# plot tree
plot(fit, uniform=TRUE,
+ main="Classification Tree ")
Error in plot.rpart(fit, uniform = TRUE, main = "Classification Tree ") :
fit is not a tree, just a root
text(fit, use.n=TRUE, all=TRUE, cex=.8)
Error in text.rpart(fit, use.n = TRUE, all = TRUE, cex = 0.8) :
fit is not a tree, just a root
what is my wrong while I applied rpart ? why it give me error with tree plot? how to fix this error Error :
fit is not a tree, just a root
You use method="class" if you are building a classification tree and method="anova" if you are building a regression tree. It looks like you have a continuous response, so you should be building a regression tree (i.e. method="anova").
You are using the RPART's default control parameters. With your data set RPART is unable to adhere to default values and create a tree (branch splitting)
rpart.control(minsplit = 20, minbucket = round(minsplit/3), cp = 0.01,
maxcompete = 4, maxsurrogate = 5, usesurrogate = 2, xval = 10,
surrogatestyle = 0, maxdepth = 30, ...)
Adjust the control parameters according to the data set.
e.g :
t <- rpart(formula = v5 ~ v1 + v2 + v3 + v4, data = x, method = anova",control =rpart.control(minsplit = 1,minbucket=1, cp=0))
But be aware this could create an over fitting decision tree.
I ran the following code with your x data frame and got a tree as shown below:
library(rpart)
library(rattle)
library(rpart.plot)
library(RColorBrewer)
fit <- rpart(v5 ~ v1+v2+v3+v4,
method="anova",
data=x,
control = rpart.control(minsplit = 6, cp = 0.01))
fancyRpartPlot(fit) #from RColorBrewer package
Note that your method should be anova as v5 is a continuous variable, and you have to override the control parameters control = rpart.control(...) to adjust the depth of the tree.
I have this data frame :
> df
Z freq proba
1 17 1 0.0033289263
2 18 4 0.0055569026
3 19 2 0.0087878028
4 20 3 0.0132023556
5 21 16 0.0188900561
6 22 12 0.0257995234
7 23 30 0.0337042731
8 24 41 0.0421963455
9 25 56 0.0507149437
10 26 65 0.0586089198
11 27 65 0.0652230449
12 28 93 0.0699913154
13 29 82 0.0725182432
14 30 94 0.0726318551
15 31 72 0.0703990113
16 32 74 0.0661024717
17 33 58 0.0601873020
18 34 66 0.0531896431
19 35 38 0.0456625487
20 36 45 0.0381117389
21 37 27 0.0309498221
22 38 17 0.0244723502
23 39 15 0.0188543771
24 40 13 0.0141629367
25 41 4 0.0103793600
26 42 1 0.0074254435
27 43 2 0.0051886582
28 45 1 0.0023658767
29 46 1 0.0015453804
30 49 2 0.0003792308
# Here are my datas :
> dput(df)
structure(list(Z = c(17, 18, 19, 20, 21, 22, 23, 24, 25, 26,
27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42,
43, 45, 46, 49), freq = c(1, 4, 2, 3, 16, 12, 30, 41, 56, 65,
65, 93, 82, 94, 72, 74, 58, 66, 38, 45, 27, 17, 15, 13, 4, 1,
2, 1, 1, 2), proba = c(0.0033289262662263, 0.00555690264007235,
0.00878780282243439, 0.0132023555702843, 0.0188900560866825,
0.0257995234198431, 0.0337042730520012, 0.0421963455163949, 0.0507149437492447,
0.0586089198012906, 0.0652230449359029, 0.0699913153996099, 0.0725182432348992,
0.0726318551493006, 0.0703990113442269, 0.0661024716831246, 0.0601873020200862,
0.0531896430528685, 0.045662548708844, 0.0381117389181843, 0.030949822142559,
0.0244723501557229, 0.01885437705459, 0.0141629366839816, 0.0103793599644779,
0.00742544354411115, 0.00518865818999788, 0.00236587669133322,
0.00154538036835848, 0.000379230768851682)), .Names = c("Z",
"freq", "proba"), row.names = c(NA, -30L), class = "data.frame")
And I want to regroup lines for which the value "freq" is < 5 with the next line, and this while the next line is < 5.
Idk if I'm clear enough so this is the output I expect :
> df2
labels effectifs pi
1 17;20 10 0.03087599
2 21 16 0.01889006
3 22 12 0.02579952
4 23 30 0.03370427
5 24 41 0.04219635
6 25 56 0.05071494
7 26 65 0.05860892
8 27 65 0.06522304
9 28 93 0.06999132
10 29 82 0.07251824
11 30 94 0.07263186
12 31 72 0.07039901
13 32 74 0.06610247
14 33 58 0.06018730
15 34 66 0.05318964
16 35 38 0.04566255
17 36 45 0.03811174
18 37 27 0.03094982
19 38 17 0.02447235
20 39 15 0.01885438
21 40 13 0.01416294
22 41;49 11 0.02728395
I did it with nested while, but I find this solution very painful and so unoptimized.
i <- 1
freqs <- c()
labels <- c()
pi <- c()
while(i < nrow(df)) {
if (df$freq[i] >= 5) {
freqs <- c(freqs, df$freq[i])
labels <- c(labels, df$Z[i])
pi <- c(pi, df$proba[i])
i <- i + 1
}
else {
count <- df$freq[i]
countPi <- df$proba[i]
k <- i
j <- i
while(df$freq[i] < 5 & i < nrow(df)) {
if (df$freq[i+1] < 5) {
count <- count + df$freq[i+1]
countPi <- countPi + df$proba[i+1]
j <- i + 1
}
i <- i + 1
}
labels <- c(labels, paste0(df$Z[k], ";", df$Z[j]))
freqs <- c(freqs, count)
pi <- c(pi, countPi)
}
}
df2 <- data.frame(labels, freqs, pi)
I'm sure there is far better, maybe with dplyr. If you have a better solution.. Thanks !
We could use the "devel" version of "data.table" as new functions are introduced (rleid). Here, we convert the "data.frame" to "data.table" (setDT(df)), create a grouping variable ("gr") based on the logical index (freq <5) using rleid. 'Z' column is 'numeric/integer' class. Create a character column ("Z1") from the "Z". Grouped by 'gr', if the "freq" is less than 5 for all the elements of that group, summarise the rows to a single row by taking the first observation of columns (.SD[1L]), remove the unwanted columns (as .SD includes "Z1" which will result in duplicate columns), append it with the "Z1" that we get from pasting the min and max value of "Z" for that group. Otherwise, leave it unchanged (else .SD). Remove the columns that we don't need by assigning it to "NULL".
library(data.table) #data.table_1.9.5
res <- setDT(df)[, gr:=rleid(freq<5)][, Z1:= as.character(Z)][,
if(all(freq<5)) c(.SD[1L][,-4, with=FALSE],
list(Z1=toString(c(min(Z), max(Z)))))
else .SD, gr][,1:2 :=NULL][]
head(res,3)
# freq proba Z1
#1: 1 0.003328926 17, 20
#2: 16 0.018890056 21
#3: 12 0.025799523 22
Since this is a dplyr question, here is a dplyr solution. First I used a grouping function in order to define the groups (similar to the rleid function in data.table). Then the summary and is fairly simple.
# grouping function
grouping <- function(condition){
# calculate runs for grouping
run <- rle((!condition) * 1:length(condition))
# revalue
run$values <- seq_along(run$values)
# invert to get grouping
inverse.rle(run)
}
# load dplyr
require(dplyr)
df %>%
mutate(group = grouping(freq<5)) %>% # add groups
group_by(group) %>% # group data
summarize(freq = sum(freq), # sum freq
proba = sum(proba), # sum proba
Z = toString(unique(range(Z)))) %>% # rename Z
mutate(group=NULL) # remove groups
## Source: local data table [22 x 3]
##
## freq proba Z
## 1 10 0.03087599 17, 20
## 2 16 0.01889006 21
## 3 12 0.02579952 22
## 4 30 0.03370427 23
## 5 41 0.04219635 24
## 6 56 0.05071494 25
## 7 65 0.05860892 26
## 8 65 0.06522304 27
## 9 93 0.06999132 28
## 10 82 0.07251824 29
## .. ... ... ...