Gompertz-Makeham parameter estimation - r

I would like estimate the parameters of the Gompert-Makeham distribution, but I haven't got a result.
I would like a method in R, like this Weibull parameter estimation code:
weibull_loglik <- function(parm){
gamma <- parm[1]
lambda <- parm[2]
loglik <- sum(dweibull(vec, shape=gamma, scale=lambda, log=TRUE))
return(-loglik)
}
weibull <- nlm(weibull_loglik,parm<-c(1,1), hessian = TRUE, iterlim=100)
weibull$estimate
c=weibull$estimate[1];b=weibull$estimate[2]
My data:
[1] 872 52 31 26 22 17 11 17 17 8 20 12 25 14 17
[16] 20 17 23 32 37 28 24 43 40 34 29 26 32 34 51
[31] 50 67 84 70 71 137 123 137 172 189 212 251 248 272 314
[46] 374 345 411 494 461 505 506 565 590 535 639 710 733 795 786
[61] 894 963 1019 1149 1185 1356 1354 1460 1622 1783 1843 2049 2262 2316 2591
[76] 2730 2972 3187 3432 3438 3959 3140 3612 3820 3478 4054 3587 3433 3150 2881
[91] 2639 2250 1850 1546 1236 966 729 532 375 256 168 107 65 39 22
[106] 12 6 3 2 1 1
summary(vec)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.0 32.0 314.0 900.9 1355.0 4054.0

It would be nice to have a reproducible example, but something like:
library(bbmle)
library(eha)
set.seed(101)
vec <- rmakeham(1000, shape = c(2,3), scale = 2)
dmwrap <- function(x, shape1, shape2, scale, log) {
res <- try(dmakeham(x, c(shape1, shape2), scale, log = log), silent = TRUE)
if (inherits(res, "try-error")) return(NA)
res
}
m1 <- mle2(y ~ dmwrap(shape1, shape2, scale),
start = list(shape1=1,shape2=1, scale=1),
data = data.frame(y = vec),
method = "Nelder-Mead"
)
Define a wrapper that (1) takes shape parameters as separate values; (2) returns NA rather than throwing an error when e.g. parameters are negative
Use Nelder-Mead rather than default BFGS for robustness
the fitdistrplus package might help too
if you're going to do a lot of this it may help to fit parameters on the log scale (i.e. use parameters logshape1, etc., and use exp(logshape1) etc. in the fitting formula)
I had to work a little harder to fit your data; I scaled the variable by 1000 (and found that I could only compute the log-likelihood; the likelihood gave an error that I didn't bother trying to track down). Unfortunately, it doesn't look like a great fit (too many small values).
x <- scan(text = "872 52 31 26 22 17 11 17 17 8 20 12 25 14 17
20 17 23 32 37 28 24 43 40 34 29 26 32 34 51
50 67 84 70 71 137 123 137 172 189 212 251 248 272 314
374 345 411 494 461 505 506 565 590 535 639 710 733 795 786
894 963 1019 1149 1185 1356 1354 1460 1622 1783 1843 2049 2262 2316 2591
2730 2972 3187 3432 3438 3959 3140 3612 3820 3478 4054 3587 3433 3150 2881
2639 2250 1850 1546 1236 966 729 532 375 256 168 107 65 39 22
12 6 3 2 1 1")
m1 <- mle2(y ~ dmwrap(shape1, shape2, scale),
start = list(shape1=1,shape2=1, scale=10000),
data = data.frame(y = x/1000),
method = "Nelder-Mead"
)
cc <- as.list(coef(m1))
png("gm.png")
hist(x,breaks = 25, freq=FALSE)
with(cc,
curve(exp(dmwrap(x/1000, shape1, shape2, scale, log = TRUE))/1000, add = TRUE)
)
dev.off()

Related

fast partial match checking in R (or Python or Julia)

I have two dataset with names and I need to compare names in both datasets. I just need to keep the union of the two datasets based on the names. However, a name is still considered 'matched' if it is part of the another name even if it is not a full match and vice versa. For example, "seb" should match to "seb", but also to "sebas". I am using str_detect(), but it is too slow. I am wondering if there is any way to speed up this process. I tried some other packages and functions, but nothing really improved the speed. I am open for any R or Python solution.
Create two dummy datasets
library(dplyr)
library(stringr)
set.seed(1)
data_set_A <- tibble(name = unique(replicate(2000, paste(sample(letters, runif(1, 3, 10), replace = T), collapse = "")))) %>%
mutate(ID_A = 1:n())
set.seed(2)
data_set_B <- tibble(name_2 = unique(replicate(2000, paste(sample(letters, runif(1, 3, 10), replace = T), collapse = "")))) %>%
mutate(ID_B = 1:n())
Test matching of full matches only
# This is almost instant
data_set_A %>%
rowwise() %>%
filter(any(name %in% data_set_B$name_2) | any(data_set_B$name_2 %in% name)) %>%
ungroup()
# A tibble: 4 x 2
name ID_A
<chr> <int>
1 vnt 112
2 fly 391
3 cug 1125
4 xgv 1280
Include partial matches (This is what I want to optimize)
This of course only gives me the subset of dataset A, but that is ok.
# This takes way too long
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, data_set_B$name_2)) | any(str_detect(data_set_B$name_2, name))) %>%
ungroup()
A tibble: 237 x 2
name ID_A
<chr> <int>
1 wknrsauuj 2
2 lyw 7
3 igwsvrzpk 16
4 zozxjpu 18
5 cgn 22
6 oqo 45
7 gkritbe 47
8 uuq 92
9 lhwfyksz 94
10 tuw 100
Fuzzyjoin method.
This also works, but is equally slow
bind_rows(
fuzzyjoin::fuzzy_inner_join(
data_set_A,
data_set_B,
by = c("name" = "name_2"),
match_fun = stringr::str_detect
) %>%
select(name, ID_A),
fuzzyjoin::fuzzy_inner_join(
data_set_B,
data_set_A,
by = c("name_2" = "name"),
match_fun = stringr::str_detect
) %>%
select(name, ID_A)
) %>%
distinct()
data.table solution
not much faster unfortunately
library(data.table)
setDT(data_set_A)
setDT(data_set_B)
data_set_A[data_set_A[, .I[any(str_detect(name, data_set_B$name_2)) |
any(str_detect(data_set_B$name_2, name))], by = .(ID_A)]$V1]
This is an [r] option aimed at reducing the number of times you are calling str_detect() (i.e., your example is slow because the function is called several thousand times; and for not using fixed() or fixed = TRUE as jpiversen already pointed out). Answer explained in comments in the code; I will try to jump on tomorrow to explain a bit more.
This should scale reasonably well and be more memory efficient than the current approach too because reduces the rowwise computations to an absolute minimum.
Benchmarks:
n = 2000
# A tibble: 4 × 13
expression min median `itr/sec` mem_alloc `gc/sec` n_itr
<bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int>
1 original() 6.67s 6.67s 0.150 31.95MB 0.300 1
2 using_fixed() 496.54ms 496.54ms 2.01 61.39MB 4.03 1
3 using_map_fixed() 493.35ms 493.35ms 2.03 60.27MB 6.08 1
4 andrew_fun() 167.78ms 167.78ms 5.96 1.59MB 0 1
n = 4000
Note: I am not sure if you need the answer to scale; but the approach of reducing the memory-intensive part does seem to do just that (although the time difference is negligible for n = 4000 for 1 iteration, IMO).
# A tibble: 4 × 13
expression min median `itr/sec` mem_alloc `gc/sec` n_itr
<bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl> <int>
1 original() 26.63s 26.63s 0.0376 122.33MB 0.150 1
2 using_fixed() 1.91s 1.91s 0.525 243.96MB 3.67 1
3 using_map_fixed() 1.87s 1.87s 0.534 236.62MB 3.20 1
4 andrew_fun() 674.36ms 674.36ms 1.48 7.59MB 0 1
Code w/ comments:
# This is so we do not retain the strings with the max number of
# characters in our pattern because we are checking with %in% already
nchar_a = nchar(data_set_A$name)
nchar_b = nchar(data_set_B$name_2)
# Creating large patterns (excluding values w/ max number of characters)
pattern_a = str_c(unique(data_set_A$name[nchar_a != max(nchar_a, na.rm = TRUE)]), collapse = "|")
pattern_b = str_c(unique(data_set_B$name_2[nchar_b != max(nchar_b, na.rm = TRUE)]), collapse = "|")
# First checking using %in%
idx_a = data_set_A$name %in% data_set_B$name_2
# Next, IDing when a(string) matches b(pattern)
idx_a[!idx_a] = str_detect(data_set_A$name[!idx_a], pattern_b)
# IDing a(pattern) matches b(string) so we do not run every row of
# a(as a pattern) against all of b
b_to_check = data_set_B$name_2[str_detect(data_set_B$name_2, pattern_a)]
# Using unmatched values of a as a pattern for the reduced set for b
idx_a[!idx_a] = vapply(data_set_A$name[!idx_a], function(name) {
any(grepl(name, b_to_check, fixed = TRUE))
}, logical(1L), USE.NAMES = FALSE)
data_set_A[idx_a, ]
# A tibble: 237 × 2
name ID_A
<chr> <int>
1 wknrsauuj 2
2 lyw 7
3 igwsvrzpk 16
4 zozxjpu 18
5 cgn 22
6 oqo 45
7 gkritbe 47
8 uuq 92
9 lhwfyksz 94
10 tuw 100
# … with 227 more rows
Reproducible R code for benchmarks
The following code is largely taken from jpiversen who provided a great answer:
library(dplyr)
library(stringr)
n = 2000
set.seed(1)
data_set_A <- tibble(name = unique(replicate(n, paste(sample(letters, runif(1, 3, 10), replace = T), collapse = "")))) %>%
mutate(ID_A = 1:n())
set.seed(2)
data_set_B <- tibble(name_2 = unique(replicate(n, paste(sample(letters, runif(1, 3, 10), replace = T), collapse = "")))) %>%
mutate(ID_B = 1:n())
original <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, data_set_B$name_2)) | any(str_detect(data_set_B$name_2, name))) %>%
ungroup()
}
using_fixed <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, fixed(data_set_B$name_2))) | any(str_detect(data_set_B$name_2, fixed(name)))) %>%
ungroup()
}
using_map_fixed <- function() {
logical_vec <- data_set_A$name %>%
purrr::map_lgl(
~any(stringr::str_detect(.x, fixed(data_set_B$name_2))) ||
any(stringr::str_detect(data_set_B$name_2, fixed(.x)))
)
data_set_A[logical_vec, ]
}
andrew_fun = function() {
nchar_a = nchar(data_set_A$name)
nchar_b = nchar(data_set_B$name_2)
pattern_a = str_c(unique(data_set_A$name[nchar_a != max(nchar_a, na.rm = TRUE)]), collapse = "|")
pattern_b = str_c(unique(data_set_B$name_2[nchar_b != max(nchar_b, na.rm = TRUE)]), collapse = "|")
idx_a = data_set_A$name %in% data_set_B$name_2
idx_a[!idx_a] = str_detect(data_set_A$name[!idx_a], pattern_b)
b_to_check = data_set_B$name_2[str_detect(data_set_B$name_2, pattern_a)]
idx_a[!idx_a] = vapply(data_set_A$name[!idx_a], function(name) {
any(grepl(name, b_to_check, fixed = TRUE))
}, logical(1L), USE.NAMES = FALSE)
data_set_A[idx_a, ]
}
bm = bench::mark(
original(),
using_fixed(),
using_map_fixed(),
andrew_fun(),
iterations = 1
)
TL;DR
The slow part is str_detect(string, pattern).
To speed it up, wrap pattern in fixed() if you got simple strings, and in coll() if you got longer, typical human text.
To get another slight speed boost, rewrite your code using purrr::map_lgl() and use this to subset your data.
Under follows examples, explanations and benchmarks.
Rewriting str_detect() using fixed() or coll()
I believe the easiest fix is to modify how str_detect() uses regex with e.g. stringr::fixed() or stringr::coll().
From ?stringr::str_detect():
Match a fixed string (i.e. by comparing only bytes), using fixed(). This is fast, but approximate. Generally, for matching human text, you'll want coll() which respects character matching rules for the specified locale.
Under is a comparison with your original code:
original <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, data_set_B$name_2)) | any(str_detect(data_set_B$name_2, name))) %>%
ungroup()
}
# Note the use of fixed()
using_fixed <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, fixed(data_set_B$name_2))) | any(str_detect(data_set_B$name_2, fixed(name)))) %>%
ungroup()
}
# Note the use of coll()
using_coll <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, coll(data_set_B$name_2))) | any(str_detect(data_set_B$name_2, coll(name)))) %>%
ungroup()
}
bm <- bench::mark(
original(),
using_fixed(),
using_coll(),
iterations = 20
)
#> Warning: Some expressions had a GC in every iteration; so filtering is disabled.
bm
#> # A tibble: 3 × 6
#> expression min median `itr/sec` mem_alloc `gc/sec`
#> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl>
#> 1 original() 6.58s 6.59s 0.152 32.4MB 0.371
#> 2 using_fixed() 501.64ms 505.51ms 1.97 61.4MB 3.94
#> 3 using_coll() 4.48s 4.5s 0.222 61.4MB 0.512
bm %>% ggplot2::autoplot(type = "violin")
#> Loading required namespace: tidyr
Created on 2022-04-02 by the reprex package (v2.0.1)
So, as we can see, wrapping your code in fixed() will make it very fast and works well on your test data. However, it might not work as well for real human text (especially non-ASCII character sets). You should test it on your original data, and use coll() as an alternative if fixed() doesn't work.
Removing rowwise()
Another step you can take to make your code a bit faster is to get rid of rowwise(). I would replace it using purrr::map_lgl() and use this logical vector to subset the dataframe. Under is an example and a benchmark against my functions defined above:
using_map_fixed <- function() {
logical_vec <- data_set_A$name %>%
purrr::map_lgl(
~any(stringr::str_detect(.x, fixed(data_set_B$name_2))) ||
any(stringr::str_detect(data_set_B$name_2, fixed(.x)))
)
data_set_A[logical_vec, ]
}
using_map_coll <- function() {
logical_vec <- data_set_A$name %>%
purrr::map_lgl(
~any(stringr::str_detect(.x, coll(data_set_B$name_2))) ||
any(stringr::str_detect(data_set_B$name_2, coll(.x)))
)
data_set_A[logical_vec, ]
}
bm <- bench::mark(
using_fixed(),
using_map_fixed(),
using_coll(),
using_map_coll(),
iterations = 20
)
#> Warning: Some expressions had a GC in every iteration; so filtering is disabled.
bm
#> # A tibble: 4 × 6
#> expression min median `itr/sec` mem_alloc `gc/sec`
#> <bch:expr> <bch:tm> <bch:tm> <dbl> <bch:byt> <dbl>
#> 1 using_fixed() 503.4ms 507.24ms 1.95 62.9MB 5.37
#> 2 using_map_fixed() 474.28ms 477.63ms 2.09 60.3MB 3.14
#> 3 using_coll() 4.49s 4.5s 0.222 61.4MB 0.489
#> 4 using_map_coll() 4.37s 4.38s 0.228 60.2MB 0.354
Created on 2022-04-02 by the reprex package (v2.0.1)
As we see, this gives another slight speed boost.
Using fixed() with data.table or fuzzyjoin
You can also use fixed() with data.table and fuzzyjoin. I have not included it here for brevity, but my benchmark shows that data.table takes about the same amount of time as my using_map_fixed() above, and fuzzyjoin takes about twice as much time.
This makes sense to me, as the slow part is str_detect(), not the method of joining/filtering, or the underlying data structure.
If you would like to use base R, the code below might be one fast option
A <- data_set_A$name
B <- data_set_B$name_2
A2B <- sapply(A, function(x) grepl(x, B, fixed = TRUE))
B2A <- sapply(B, function(x) grepl(x, A, fixed = TRUE))
idx <- which(t(A2B) | B2A, arr.ind = TRUE)
res <- cbind(data_set_A[idx[, 1], ], data_set_B[idx[, 2], ])
which gives
> res
name ID_A name_2 ID_B
1 arh 1234 pimoarhd 8
2 qtj 720 aqtj 23
3 szcympsn 142 cym 43
4 cymvubnxg 245 cym 43
5 dppvtcymq 355 cym 43
6 kzi 690 kzii 48
7 eyajqchkn 498 chk 53
8 upfzh 522 upf 61
9 ioa 1852 ioadr 63
10 lya 1349 ibelyalvh 64
11 honod 504 ono 71
12 zozxjpu 18 zoz 72
13 jcz 914 cdjczpqg 88
14 ailmjf 623 ilm 99
15 upoux 609 oux 104
16 pouxifvp 1466 oux 104
17 mvob 516 vob 106
18 nqtotvhhm 1088 otv 115
19 wom 202 womtglapx 117
20 qkc 756 dqkcfqpps 118
21 qtl 600 ivqtlymzr 126
22 qqi 1605 owfsqqiyu 153
23 fmjalirze 1470 ali 172
24 ibwfwkyp 1588 fwk 175
25 iat 1258 iatjeg 185
26 osm 253 nviiqosm 199
27 wpj 373 wpjeb 204
28 hahx 515 ahx 213
29 keahxa 1565 ahx 213
30 psf 359 qnpsfo 223
31 saq 1859 saqhu 227
32 cvmkwtx 714 cvm 228
33 ilw 389 pyilwj 231
34 ohwysv 1590 ysv 237
35 utrl 698 trl 244
36 dmttrlcpj 1267 trl 244
37 cpv 236 btcpvmoc 247
38 uto 1047 utoi 257
39 yngunekl 1978 ekl 258
40 vceko 625 vce 265
41 fir 1934 firgk 278
42 qvd 983 eqvdfi 287
43 fir 1934 zwwefir 291
44 idvfkevdf 1380 vdf 312
45 qwdo 1921 qwd 322
46 kam 1205 tlkam 327
47 lck 488 clckjkyzn 329
48 gmspwckw 1015 msp 359
49 ynouuwqtz 1576 nou 360
50 tty 1209 bttyvt 361
51 vkc 999 fmrvkcl 366
52 ipw 1918 fipwjomdu 388
53 zdv 261 zdvkut 410
54 vku 1137 zdvkut 410
55 doby 246 oby 411
56 hycvuupgy 141 uup 421
57 uwlb 1249 wlb 431
58 auj 1452 lcmnauj 444
59 rwd 1667 ukwrwdczs 479
60 ylsihqqor 1290 ihq 483
61 feo 1649 feorvxbm 485
62 zff 755 dohzffujm 499
63 mqutujepu 904 epu 507
64 uiepu 1308 epu 507
65 vahepuk 1434 epu 507
66 cug 1125 accugl 509
67 fir 1934 firwe 517
68 dia 1599 dialeddd 527
69 temiwd 1725 tem 531
70 svofivl 1177 svo 545
71 flm 657 aflm 546
72 vnt 112 vnt 551
73 bhmoskrz 426 osk 558
74 wev 728 shemuwev 569
75 hzpi 1586 hzp 579
76 gvi 1064 mkgvivlfe 582
77 fjb 1398 vkfjbxnjl 589
78 qin 1013 qinp 593
79 ecn 1342 ecnzre 598
80 zre 1610 ecnzre 598
81 xvr 772 dpxvrfmo 623
82 tqr 1419 tqrmztdm 624
83 zmwnf 1571 mwn 626
84 ypil 1787 pil 630
85 mnxlqgfh 1132 nxl 643
86 gse 1563 gseice 646
87 ygk 1309 ygkqrk 655
88 fgm 933 vzfgmy 663
89 rlupd 977 upd 666
90 mcupdkuiy 1307 upd 666
91 fly 391 fly 669
92 vbkko 1603 kko 678
93 uvrew 465 rew 680
94 hgbhngwvd 901 wvd 690
95 wvdjprmo 1432 wvd 690
96 cgn 22 cgnd 698
97 dngnjv 967 njv 700
98 psqs 841 sqs 720
99 ywv 1180 ptywvlgc 730
100 ypil 1787 ypi 734
101 rwd 1667 srserwd 737
102 jqydasl 1294 jqy 742
103 ckujmc 717 ujm 751
104 dfzxta 662 xta 775
105 bjb 1562 jabjbei 779
106 adwknpll 1242 npl 780
107 kdv 1327 xhkdvqo 789
108 ghj 174 oghj 801
109 lhwfyksz 94 lhw 811
110 nwrrnlhw 929 lhw 811
111 xlhwm 1720 lhw 811
112 ncc 1602 wurhxnccn 814
113 jdslrf 1094 dsl 835
114 ktmw 1738 tmw 844
115 igwsvrzpk 16 gws 856
116 kug 591 pkugls 857
117 befgcpedr 339 fgc 862
118 ojf 1397 ojfpnkla 863
119 gyl 1203 gylxeqzw 872
120 ugcbb 1727 ugc 876
121 arh 1234 karhwhg 878
122 amm 458 ammqdc 883
123 azazryje 636 zaz 900
124 wczazw 1887 zaz 900
125 gkritbe 47 ritb 915
126 vku 1137 yjvkuxued 929
127 rnh 1633 kvyrnhugu 937
128 mzh 1135 xllmwmzhn 940
129 cug 1125 cug 960
130 xgv 1280 xgv 962
131 xusxgv 1436 xgv 962
132 umc 351 lwumcmvoo 980
133 zlb 1900 nkyazlb 991
134 llfkalao 1049 llf 1002
135 sflpbht 991 lpb 1048
136 rairmmcl 442 mmc 1087
137 mmckoln 780 mmc 1087
138 gfxmmcgb 1814 mmc 1087
139 aoj 402 taojlgp 1089
140 mypvzhp 121 ypv 1095
141 moctwaypv 611 ypv 1095
142 rngedn 306 ged 1106
143 djshecy 1408 ecy 1108
144 rairmmcl 442 rmm 1117
145 gzua 1594 zua 1124
146 ytj 416 yytj 1140
147 ubt 300 hubtcfr 1141
148 gqg 1854 ogqgsjqc 1144
149 tfg 1204 xiutfgru 1145
150 avrq 741 avr 1147
151 ytkpvss 440 tkp 1149
152 kug 591 yxsjkug 1176
153 vix 1846 vixsmn 1187
154 qtl 600 qtljkxz 1188
155 lgr 494 dlgrco 1189
156 ryg 864 xlmtryg 1203
157 yskvkxwj 1547 kvk 1205
158 kxhee 1795 xhe 1222
159 hzbcjs 1493 cjs 1224
160 kbi 270 itxlwkbi 1225
161 gdymcam 806 ymca 1232
162 tqr 1419 rxtqrdtl 1236
163 yyz 215 yyzw 1242
164 jyx 1735 mljjyxu 1248
165 aai 1928 umkpaaiwo 1254
166 dsd 1122 dndsdova 1257
167 tor 744 etor 1270
168 vhcyznp 1296 yzn 1278
169 xlc 1947 odxlcjwj 1280
170 mlm 1629 aomlmgtq 1303
171 owm 239 owmugb 1304
172 ynezwaml 507 nez 1308
173 jls 695 jlsve 1325
174 dvm 879 dvmv 1339
175 vsgx 944 dqpihvsgx 1352
176 wfo 768 wfokpjois 1354
177 tltbkinat 1986 nat 1362
178 gyl 1203 gylqte 1363
179 ngg 735 bsnggqbjd 1366
180 fkq 345 jdfkqf 1368
181 ojf 1397 ojfpgfga 1382
182 dqgd 1623 prqbndqgd 1398
183 siu 827 siuypucup 1412
184 yinsoivfd 1895 yin 1414
185 esm 1834 sesmeepz 1417
186 umc 351 umcj 1432
187 wny 866 wnyxamguw 1443
188 ujbhtvnin 399 vni 1444
189 dbq 630 bdbqq 1452
190 ebn 1405 ebngddw 1461
191 zcj 704 rbtjzcjod 1465
192 avn 500 avnspxv 1468
193 vkk 567 hvkk 1477
194 hmm 1441 bgjhmmthz 1483
195 aguakz 614 guak 1487
196 hycvuupgy 141 pgy 1493
197 tizpgymz 280 pgy 1493
198 guk 571 cncxdguk 1502
199 zyw 281 nzywuqs 1504
200 jnz 1558 rxdxsjnzw 1510
201 uuq 92 nxuuqtj 1514
202 qtj 720 nxuuqtj 1514
203 vkk 567 xpbpvkkdc 1518
204 iaa 460 sjiaa 1525
205 txsgmynng 1019 xsg 1526
206 yjvtwc 1107 jvt 1529
207 lnk 1113 hylnknwy 1546
208 szd 635 woszdm 1557
209 osm 253 sosmdp 1567
210 nbd 1067 nbdmmg 1570
211 mmg 1305 nbdmmg 1570
212 wqdsatbd 1536 sat 1585
213 sdlypo 1527 sdl 1596
214 inkynog 288 inky 1600
215 hpwoeclfy 1321 clf 1601
216 wodyqwqf 679 dyq 1603
217 lyw 7 xnalywyuw 1607
218 njm 1825 vjlnjmns 1617
219 njytqhaut 428 qha 1620
220 ilw 389 rilwbk 1647
221 oqo 45 ixoqowkpg 1650
222 odcbcvaun 1386 bcv 1652
223 mastn 434 stn 1662
224 xebhdssit 1091 xeb 1663
225 nmy 782 nmyxj 1671
226 fsqvgdw 673 gdw 1676
227 mwwczhs 482 wcz 1679
228 wczazw 1887 wcz 1679
229 anmryzm 915 ryz 1698
230 rteh 523 rte 1708
231 mlwrguae 817 lwr 1709
232 mbu 819 xpsuqmbuf 1729
233 mmckoln 780 cko 1733
234 lxpg 798 lxp 1734
235 ane 370 vxnanehvk 1746
236 tty 1209 vbttyozui 1752
237 igncdgyjx 332 ign 1753
238 ndignk 621 ign 1753
239 nmy 782 ivnmyba 1780
240 wknrsauuj 2 rsa 1799
241 tgd 165 qtgdidlf 1803
242 iaa 460 yziaazxto 1833
243 xto 1245 yziaazxto 1833
244 zff 755 dpzfft 1857
245 jyx 1735 jwjyxphe 1873
246 ytj 416 eytj 1881
247 lcggwonk 1596 onk 1882
248 zdv 261 zdvxfz 1889
249 xhskcb 417 kcb 1890
250 mrikqkcb 770 kcb 1890
251 psvxqnsap 1352 psv 1898
252 udjswzb 411 jsw 1900
253 rpfjswy 1840 jsw 1900
254 bjaywiso 1677 ayw 1902
255 zfli 130 fli 1906
256 vazx 1215 itvazxw 1918
257 tuw 100 tuwywtbwd 1921
258 vle 1437 ebvleaovm 1937
259 znycsygd 1757 nyc 1944
260 ynezwaml 507 ezw 1952
261 tseezwf 1276 ezw 1952
262 ezwzyfudo 1690 ezw 1952
263 oudiky 1503 dik 1964
264 dikjn 1615 dik 1964
265 oms 106 wpomsudi 1977
266 hhp 1864 hhpkm 1983
Benchmarking
It seems this base R option is slightly slower than #Andrew's approach.
TIC <- function() {
A <- data_set_A$name
B <- data_set_B$name_2
A2B <- sapply(A, function(x) grepl(x, B, fixed = TRUE))
B2A <- sapply(B, function(x) grepl(x, A, fixed = TRUE))
idx <- which(t(A2B) | B2A, arr.ind = TRUE)
cbind(data_set_A[idx[, 1], ], data_set_B[idx[, 2], ])
# data_set_A[unique(idx[, 1]), ]
}
jpiversen_fixed <- function() {
data_set_A %>%
rowwise() %>%
filter(any(str_detect(name, fixed(data_set_B$name_2))) | any(str_detect(data_set_B$name_2, fixed(name)))) %>%
ungroup()
}
andrew <- function() {
nchar_a <- nchar(data_set_A$name)
nchar_b <- nchar(data_set_B$name_2)
pattern_a <- str_c(unique(data_set_A$name[nchar_a != max(nchar_a, na.rm = TRUE)]), collapse = "|")
pattern_b <- str_c(unique(data_set_B$name_2[nchar_b != max(nchar_b, na.rm = TRUE)]), collapse = "|")
idx_a <- data_set_A$name %in% data_set_B$name_2
idx_a[!idx_a] <- str_detect(data_set_A$name[!idx_a], pattern_b)
b_to_check <- data_set_B$name_2[str_detect(data_set_B$name_2, pattern_a)]
idx_a[!idx_a] <- vapply(data_set_A$name[!idx_a], function(name) {
any(grepl(name, b_to_check, fixed = TRUE))
}, logical(1L), USE.NAMES = FALSE)
data_set_A[idx_a, ]
}
bm <- microbenchmark(
TIC(),
jpiversen_fixed(),
andrew(),
times = 20
)
shows that
> bm
Unit: milliseconds
expr min lq mean median uq max
TIC() 423.8410 441.3574 492.6091 478.2596 549.2376 611.3841
jpiversen_fixed() 1354.8954 1373.9502 1447.8649 1395.6766 1459.7058 1842.2574
andrew() 329.4821 335.3388 345.8890 341.4758 354.1298 381.6872
neval
20
20
20

Divide paired matching columns

I have a data.frame df with matching columns that are also paired. The matching columns are defined in the factor patient. I would like to devide the matching columns by each other. Any suggestions how to do this?
I tried this, but this does not take the pairing from patient into account.
m1 <- m1[sort(colnames(df)]
m1_g <- m1[,grep("^n",colnames(df))]
m1_r <- m1[,grep("^t",colnames(df))]
m1_new <- m1_g/m1_r
m1_new
head(df)
na-008 ta-008 nc012 tb012 na020 na-018 ta-018 na020 tc020 tc093 nc093
hsa-let-7b-5p_TGAGGTAGTAGGTTGTGT 56 311 137 242 23 96 113 106 41 114
hsa-let-7b-5p_TGAGGTAGTAGGTTGTGTGG 208 656 350 713 49 476 183 246 157 306
hsa-let-7b-5p_TGAGGTAGTAGGTTGTGTGGT 631 1978 1531 2470 216 1906 732 850 665 909
hsa-let-7b-5p_TGAGGTAGTAGGTTGTGTGGTT 2760 8159 6067 9367 622 4228 2931 3031 2895 2974
hsa-let-7b-5p_TGAGGTAGTAGGTTGTGTGGTTT 1698 4105 3737 3729 219 1510 1697 1643 1527 1536
> head(patient)
$`008`
[1] "na-008" "ta-008"
$`012`
[1] "nc012" "tb012"
$`018`
[1] "na-018" "ta-018"
$`020`
[1] "na020" "tc020"
$`045`
[1] "nb045" "tc045"
$`080`
[1] "nb-080" "ta-080"

Increment every nth range R

I am writing R code where there's a vector 'x' which contains values 1 to 100 and I want to create another vector 'y' which subsets a range of values at every nth range. I'm sure I can use the rep() and seq() but I can't figure out the code to get what I need. Here's what the output should look like
x <- 1:100
y <- 1 2 3 11 12 13 21 22 23 31 32 33 41 42 43 51 52 53 61 62 63 71 72 73 81 82 83 91 92 93
So if I was to do have a vector x <- 1001:1100, x[y] should return:
1001 1002 1003 1011 1012 1013 1021 1022 1023 1031 1032 1033 1041 1042 1043...etc
Any ideas?
You could use grepl for that:
x <- 1001:1100
y <- grepl("[1-3]$", x)
x[y]
# [1] 1001 1002 1003 1011 1012 1013 1021 1022 1023 1031 1032 1033 1041 1042 1043 1051 1052
#[18] 1053 1061 1062 1063 1071 1072 1073 1081 1082 1083 1091 1092 1093
It simply checks for each element of x whether the last digit is in the range of 1, 2 or 3 and if so, it returns TRUE, otherwise FALSE. This logical index is then used to subset x.
In case your objective is not to subset elements ending in 1,2 or 3 but instead, to always subset 3 elements, then leave out 7, and then subset 3 again etc... you could do:
x <- 1001:1100
y <- rep(c(TRUE, FALSE), c(3, 7))
x[y]
# [1] 1001 1002 1003 1011 1012 1013 1021 1022 1023 1031 1032 1033 1041 1042 1043 1051 1052
#[18] 1053 1061 1062 1063 1071 1072 1073 1081 1082 1083 1091 1092 1093
In this case, vector y which is again logical, is recycled - note that length(x) should be divisible by length(y) for this to work properly.
For fun, With outer:
x <- 1001:1100
y <- as.vector(outer(1:3, seq(0, length(x)-10, 10), "+"))
x[y]
# [1] 1001 1002 1003 1011 1012 1013 1021 1022 1023 1031 1032 1033 1041 1042 1043
# [16] 1051 1052 1053 1061 1062 1063 1071 1072 1073 1081 1082 1083 1091 1092 1093
Probably this may help you:
x <- 1:100
y <- as.integer()
for(i in seq(1, length(x), 10)) {
y <- append(y, c(x[i], x[i+1], x[i+2]))
}
Hm. This started out as fun, but now I happen to like it since it is constructed in basically the same way the author of the question put it:
> do.call("c",lapply(0:5,function(X) 1:3+10*X))
[1] 1 2 3 11 12 13 21 22 23 31 32 33 41 42 43 51 52 53

interpreting dates from Auto arima model

The following is my code,
auto<-auto.arima(x)
auto_for<-forecast(auto,h=30)
> auto_for$x
Time Series:
Start = 1
End = 74
Frequency = 1
[1] 151 151 151 151 151 219 465 465 465 465 465 743 743 743 743 743 743 743 743 743 743 743
[23] 743 743 743 743 743 743 743 829 829 829 829 829 829 1004 1004 1004 1424 1424 1424 1822 1941 1941
[45] 1941 1941 1941 1941 1941 2076 2076 2252 2252 2252 2252 2252 2252 2252 2252 2252 2252 2252 2252 2940 2940 2940
[67] 2940 2940 3134 3134 3134 3207 3207 3465
> auto_for
Point Forecast Lo 80 Hi 80 Lo 95 Hi 95
75 3510.397 3359.577 3661.217 3279.738 3741.056
76 3555.795 3342.503 3769.086 3229.594 3881.996
77 3601.192 3339.964 3862.419 3201.679 4000.705
78 3646.589 3344.949 3948.229 3185.271 4107.907
79 3691.986 3354.743 4029.230 3176.217 4207.755
80 3737.384 3367.952 4106.815 3172.387 4302.380
81 3782.781 3383.749 4181.812 3172.515 4393.047
82 3828.178 3401.595 4254.761 3175.776 4480.580
83 3873.575 3421.116 4326.035 3181.599 4565.552
84 3918.973 3442.039 4395.907 3189.565 4648.380
85 3964.370 3464.157 4464.582 3199.361 4729.379
86 4009.767 3487.312 4532.222 3210.741 4808.793
87 4055.164 3511.376 4598.953 3223.512 4886.817
88 4100.562 3536.246 4664.878 3237.515 4963.608
89 4145.959 3561.836 4730.081 3252.621 5039.297
90 4191.356 3588.077 4794.635 3268.720 5113.992
91 4236.753 3614.908 4858.599 3285.722 5187.785
I have the forecasted value, but I am not able to get the dates from the model. The dates are not present in the graph either and it has changed from 0 to 91, instead of my actual dates. I have used xts variable at the starting.
Update:
> a<-ts(ana)
> a
Time Series:
Start = 1
End = 68
Frequency = 1
final.day final.cumsum135
1 16535 318
2 16536 318
3 16537 318
4 16538 318
5 16539 318
6 16540 318
7 16541 318
8 16542 318
9 16543 318
10 16544 318
11 16545 318
12 16546 318
13 16547 318
14 16548 318
15 16549 318
16 16550 318
17 16551 318
18 16552 318
19 16553 318
20 16554 318
21 16555 318
22 16556 318
23 16557 318
24 16558 318
25 16559 318
26 16560 369
27 16561 369
28 16562 369
29 16563 369
30 16564 369
31 16565 369
32 16566 369
33 16567 369
34 16568 369
35 16569 369
> auto<-arima(a)
Error in arima(a) : only implemented for univariate time series
Is there any way I can get back the dates here?
Whit daily series, sometimes fitted and forecast "lost" dates. You could get dates by hand, using index:
y=x # x is your xts series
n=length(y)
model_a1 <- auto.arima(y)
# the plot
plot(x=1:n,y,xaxt="n",xlab="")
axis(1,at=seq(1,n,length.out=20),labels=index(y)[seq(1,n,length.out=20)],
las=2,cex.axis=.5)
lines(fitted(model_a1), col = 2)
#the forecast
auto_for<-forecast(model_a1,h=30)
fcs=xts(auto_for$mean,seq.Date(as.Date(index(y)[n]),by=1,length.out=30))
fcs

Why does my detrending not result in a different semivariogram in R?

I have a series of samples, which I wish to construct a variogram model of, and eventual kriging model. First, I neeed to detrend the data, as shown below:
samples
x y z
1 180 1180 2.763441
2 180 240 -2.000000
3 380 1840 1.720087
4 720 80 4.056754
5 860 800 4.361503
6 620 1360 4.737717
7 980 1920 4.352956
8 1680 260 4.568255
9 1520 800 5.025272
10 1100 1220 4.693432
11 800 1460 2.470927
12 360 1900 1.455169
13 700 1760 2.894159
14 720 1540 2.115742
15 660 1480 1.749017
16 540 1680 3.291592
17 260 1280 2.962401
18 440 1640 2.422442
19 280 1260 2.966076
20 580 1580 3.178913
21 600 1220 3.752786
22 240 1700 1.748011
23 480 1440 3.106302
24 740 1880 4.827699
25 760 1320 3.603621
26 1560 1640 5.410076
27 1960 1980 6.145778
28 1520 1620 5.499064
29 1900 1820 5.316121
30 1780 1580 5.318344
31 100 740 2.019103
32 180 760 2.353693
33 140 200 1.714856
34 380 720 3.526107
35 240 580 3.075283
36 260 600 3.329397
37 340 360 3.188613
38 280 680 2.626241
39 420 700 3.211163
40 500 240 2.960805
41 460 280 3.171664
42 480 300 2.828883
43 400 640 3.227938
44 440 480 2.420358
45 300 560 4.021187
46 1380 220 5.364264
47 1500 740 5.344526
48 1240 380 4.632060
49 1420 360 4.012537
50 1280 800 4.122139
51 1400 600 5.033020
52 1300 640 4.215308
53 1460 200 5.116025
54 1220 440 4.550290
55 1200 520 3.788613
56 1540 340 5.772432
57 1520 660 5.656598
58 1480 260 5.423685
59 1360 780 4.728220
60 1260 240 3.683696
print(mean(samples$z))
h <- gstat(formula=z~1, locations=~x+y, data=samples)
samples.vgm <- variogram(h)
plot(samples.vgm,main='Variogram of Samples NOT detrended')
z = samples$z
x = samples$x
y = samples$y
trend <- lm(z~x+y)
c = trend$coefficients[[1]]
a = trend$coefficients[[2]]
b = trend$coefficients[[3]]
Xs <- c()
Ys <- c()
Zs <- c()
print('started the loop')
for (i in 1:nrow(samples)){
i = samples[i,]
x=i$x
y=i$y
z=i$z
z_prime = z - (a*x+b*y+c)
Xs <- c(Xs,x)
Ys <- c(Ys,y)
Zs <- c(Zs,z_prime)
}
sampled <- data.frame(Xs,Ys,Zs)
print(sampled)
print('the length of sampled is')
print(length(sampled[[1]]))
print(levelplot(Zs~Xs+Ys,sampled))
x <- seq(0,2000,by=20)
y <- seq(0,2000,by=20)
pred.grid <- data.frame(x=rep(x,times=length(y)),y=rep(y,each=length(x)))
g <- gstat(formula=Zs~1, locations=~Xs+Ys, data=sampled)
sampled.vgm <- variogram(g)
plot(sampled.vgm,main='Variogram of Samples hopefully detrended')
The problem is that the plot of detrended variogram (i.e. the variogram g above) looks exactly the same as the variogram h also above, which is NOT detrended. any reason why this happens??
The data is clearly different, The mean of the values in the detrended version is 0, as expected, but the non-detrended version the mean is around 3.556, also as expected.
Is there something I'm not catching here?
Not sure this question belongs here, since I think the issue is conceptual, not related to your code. I'm new, though, so I'll just go ahead and give you some quick feedback.
The variogram plots the variance (or semi-variance technically) of your data within a given spatial lag. When you apply a linear transformation to your data, I don't believe you'll alter the variance, and so you shouldn't see different parameters come out of the variogram model. Instead, your kriged surface just takes on a different mean value.
p.s. It would be helpful to make the code something that anyone can copy and paste -- e.g., include coded test data.

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