英文:
Sorting Data with custom sequence in R
问题
我有一个问题，需要按文件名对数据库中的数据进行排序。
首先，我有一个包含20个netcdf数据的文件列表。这是数据列表：
file_all <- list.files(pattern=glob2rx("*.nc"))> file_all[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-jja-tas-cru.nc"[3] "bs-cnrmaladin-mam-tas-cru.nc" "bs-cnrmaladin-son-tas-cru.nc"[5] "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-jja-tas-cru.nc"[7] "bs-ecearthcclm-mam-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc"[9] "bs-hadgemhirham-djf-tas-cru.nc" "bs-hadgemhirham-jja-tas-cru.nc"[11] "bs-hadgemhirham-mam-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-jja-tas-cru.nc"[15] "bs-mpiracmo-mam-tas-cru.nc" "bs-mpiracmo-son-tas-cru.nc"[17] "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-jja-tas-cru.nc"[19] "bs-noresmremo-mam-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
我需要首先对它进行排序，因为我需要顺序的数据。我想要的顺序是：
djf -> mam -> jja -> son
这是我期望的输出顺序：
[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-mam-tas-cru.nc"[3] "bs-cnrmaladin-jja-tas-cru.nc" "bs-cnrmaladin-son-tas-cru.nc"[5] "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-mam-tas-cru.nc"[7] "bs-ecearthcclm-jja-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc"[9] "bs-hadgemhirham-djf-tas-cru.nc" "bs-hadgemhirham-mam-tas-cru.nc"[11] "bs-hadgemhirham-jja-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-mam-tas-cru.nc"[15] "bs-mpiracmo-jja-tas-cru.nc" "bs-mpiracmo-son-tas-cru.nc"[17] "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-mam-tas-cru.nc"[19] "bs-noresmremo-jja-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
如果有人知道如何按照这个顺序排序，请帮助我。
提前感谢。
英文:
I have a problem with sorting data base on the name of the files.
First I have list file containing 20 netcdf data. This is the data list :
file_all <- list.files(pattern=glob2rx("*.nc"))> file_all[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-jja-tas-cru.nc"[3] "bs-cnrmaladin-mam-tas-cru.nc" "bs-cnrmaladin-son-tas-cru.nc"[5] "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-jja-tas-cru.nc"[7] "bs-ecearthcclm-mam-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc"[9] "bs-hadgemhirham-djf-tas-cru.nc" "bs-hadgemhirham-jja-tas-cru.nc"[11] "bs-hadgemhirham-mam-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-jja-tas-cru.nc"[15] "bs-mpiracmo-mam-tas-cru.nc" "bs-mpiracmo-son-tas-cru.nc"[17] "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-jja-tas-cru.nc"[19] "bs-noresmremo-mam-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
I need to sort it first because I need sequential data. The order that I want is:
djf -> mam -> jja -> son
This is my expected output order:
[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-mam-tas-cru.nc"[3] "bs-cnrmaladin-jja-tas-cru.nc" "bs-cnrmaladin-son-tas-cru.nc"[5] "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-mam-tas-cru.nc"[7] "bs-ecearthcclm-jja-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc"[9] "bs-hadgemhirham-djf-tas-cru.nc" "bs-hadgemhirham-mam-tas-cru.nc"[11] "bs-hadgemhirham-jja-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-mam-tas-cru.nc"[15] "bs-mpiracmo-jja-tas-cru.nc" "bs-mpiracmo-son-tas-cru.nc"[17] "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-mam-tas-cru.nc"[19] "bs-noresmremo-jja-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
If anyone know how to sort with this sequence please help me.
Thanks in advance
答案1
得分: 2
以下是您要翻译的代码部分：
你可以提取字符串的这一部分并与你的顺序进行匹配，即
my_order <- c("djf", "mam", "jja", "son")i1 <- gsub(".*-(.*)-tas.*", "\", file_all)i2 <- gsub("bs-([^-]*)-.*", "\", file_all)file_all[order(i2, match(i1, my_order))]
一些替代方法包括：
使用 strsplit。这可以提高效率，因为它不使用正则表达式。
split_files <- strsplit(strings, "-")i3 <- sapply(split_files, "[[", 3)i4 <- sapply(split_files, "[[", 2)file_all[order(i4, match(i3, my_order))]
英文:
You can extract that part of the string and match it with your order, i.e.
my_order <- c("djf", "mam", "jja", "son")i1 <- gsub(".*-(.*)-tas.*", "\", file_all)i2 <- gsub("bs-([^-]*)-.*", "\", file_all)file_all[order(i2, match(i1, my_order))][1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-mam-tas-cru.nc" "bs-cnrmaladin-jja-tas-cru.nc" "bs-cnrmaladin-son-tas-cru.nc"[5] "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-mam-tas-cru.nc" "bs-ecearthcclm-jja-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc"[9] "bs-hadgemhirham-djf-tas-cru.nc" "bs-hadgemhirham-mam-tas-cru.nc" "bs-hadgemhirham-jja-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-mam-tas-cru.nc" "bs-mpiracmo-jja-tas-cru.nc" "bs-mpiracmo-son-tas-cru.nc"[17] "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-mam-tas-cru.nc" "bs-noresmremo-jja-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
A few alternatives would be:
To use strsplit. This could improve efficiency as it doesn't use regular expressions.
split_files <- strsplit(strings, "-")i3 <- sapply(split_files, "[[", 3)i4 <- sapply(split_files, "[[", 2)file_all[order(i4, match(i3, my_order))]
答案2
得分: 1
A little bit verbose, but you could use a tidyverse based solution:
library(dplyr)library(tidyr)file_all %>%data.frame(model = .) %>%separate_wider_delim(model, delim = "-", names_sep = "", cols_remove = FALSE) %>%mutate(model3 = match(model3, my_order)) %>%arrange(model2, model3) %>%pull(modelmodel)
This returns:
[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-mam-tas-cru.nc" "bs-cnrmaladin-jja-tas-cru.nc"[4] "bs-cnrmaladin-son-tas-cru.nc" "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-mam-tas-cru.nc"[7] "bs-ecearthcclm-jja-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc" "bs-hadgemhirham-djf-tas-cru.nc"[10] "bs-hadgemhirham-mam-tas-cru.nc" "bs-hadgemhirham-jja-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-mam-tas-cru.nc" "bs-mpiracmo-jja-tas-cru.nc"[16] "bs-mpiracmo-son-tas-cru.nc" "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-mam-tas-cru.nc"[19] "bs-noresmremo-jja-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
Data:
file_all <- c("bs-cnrmaladin-djf-tas-cru.nc", "bs-cnrmaladin-jja-tas-cru.nc","bs-cnrmaladin-mam-tas-cru.nc", "bs-cnrmaladin-son-tas-cru.nc","bs-ecearthcclm-djf-tas-cru.nc", "bs-ecearthcclm-jja-tas-cru.nc","bs-ecearthcclm-mam-tas-cru.nc", "bs-ecearthcclm-son-tas-cru.nc","bs-hadgemhirham-djf-tas-cru.nc", "bs-hadgemhirham-jja-tas-cru.nc","bs-hadgemhirham-mam-tas-cru.nc", "bs-hadgemhirham-son-tas-cru.nc","bs-mpiracmo-djf-tas-cru.nc", "bs-mpiracmo-jja-tas-cru.nc","bs-mpiracmo-mam-tas-cru.nc", "bs-mpiracmo-son-tas-cru.nc","bs-noresmremo-djf-tas-cru.nc", "bs-noresmremo-jja-tas-cru.nc","bs-noresmremo-mam-tas-cru.nc", "bs-noresmremo-son-tas-cru.nc")my_order <- c("djf", "mam", "jja", "son")
Benchmark:
library(microbenchmark)my_order <- c("djf", "mam", "jja", "son")my_files_2 <- sample(file_all, 100000, replace = TRUE)microbenchmark(sotos = my_files_2[order(match(gsub(".*-(.*)-tas.*", "\", my_files_2), my_order))],sotos_2 = { split_files <- strsplit(my_files_2, "-")i3 <- unlist(lapply(split_files, "[[", 3))i4 <- unlist(lapply(split_files, "[[", 2))my_files_2[order(i4, match(i3, my_order))] },sotos_3 = { split_files <- strsplit(my_files_2, "-", fixed = TRUE)i3 <- unlist(lapply(split_files, "[[", 3))i4 <- unlist(lapply(split_files, "[[", 2))my_files_2[order(i4, match(i3, my_order))] },alt = my_files_2[unlist(lapply(my_order, \(n) which(grepl(n, my_files_2))))],tidyverse = my_files_2 %>%data.frame(model = .) %>%separate_wider_delim(model, delim = "-", names_sep = "", cols_remove = FALSE) %>%mutate(model3 = match(model3, my_order)) %>%arrange(model2, model3) %>%pull(modelmodel))
Benchmark results based on the size of my_list_2 are also provided.
英文:
A little bit verbose, but you could use a tidyverse based solution:
library(dplyr)library(tidyr)file_all %>%data.frame(model = .) %>%separate_wider_delim(model, delim = "-", names_sep = "", cols_remove = FALSE) %>%mutate(model3 = match(model3, my_order)) %>%arrange(model2, model3) %>%pull(modelmodel)
This returns
[1] "bs-cnrmaladin-djf-tas-cru.nc" "bs-cnrmaladin-mam-tas-cru.nc" "bs-cnrmaladin-jja-tas-cru.nc"[4] "bs-cnrmaladin-son-tas-cru.nc" "bs-ecearthcclm-djf-tas-cru.nc" "bs-ecearthcclm-mam-tas-cru.nc"[7] "bs-ecearthcclm-jja-tas-cru.nc" "bs-ecearthcclm-son-tas-cru.nc" "bs-hadgemhirham-djf-tas-cru.nc"[10] "bs-hadgemhirham-mam-tas-cru.nc" "bs-hadgemhirham-jja-tas-cru.nc" "bs-hadgemhirham-son-tas-cru.nc"[13] "bs-mpiracmo-djf-tas-cru.nc" "bs-mpiracmo-mam-tas-cru.nc" "bs-mpiracmo-jja-tas-cru.nc"[16] "bs-mpiracmo-son-tas-cru.nc" "bs-noresmremo-djf-tas-cru.nc" "bs-noresmremo-mam-tas-cru.nc"[19] "bs-noresmremo-jja-tas-cru.nc" "bs-noresmremo-son-tas-cru.nc"
Data
file_all <- c("bs-cnrmaladin-djf-tas-cru.nc" ,"bs-cnrmaladin-jja-tas-cru.nc" ,"bs-cnrmaladin-mam-tas-cru.nc" ,"bs-cnrmaladin-son-tas-cru.nc" ,"bs-ecearthcclm-djf-tas-cru.nc" ,"bs-ecearthcclm-jja-tas-cru.nc" ,"bs-ecearthcclm-mam-tas-cru.nc" ,"bs-ecearthcclm-son-tas-cru.nc" ,"bs-hadgemhirham-djf-tas-cru.nc" ,"bs-hadgemhirham-jja-tas-cru.nc","bs-hadgemhirham-mam-tas-cru.nc" ,"bs-hadgemhirham-son-tas-cru.nc","bs-mpiracmo-djf-tas-cru.nc" ,"bs-mpiracmo-jja-tas-cru.nc" ,"bs-mpiracmo-mam-tas-cru.nc" ,"bs-mpiracmo-son-tas-cru.nc" ,"bs-noresmremo-djf-tas-cru.nc" ,"bs-noresmremo-jja-tas-cru.nc" ,"bs-noresmremo-mam-tas-cru.nc" ,"bs-noresmremo-son-tas-cru.nc" )my_order <- c("djf", "mam", "jja", "son")
Benchmark
Since I was curious I did a small benchmark of @Sotos answer vs. a variation of this answer and the tidyverse based solution:
library(microbenchmark)my_order <- c("djf", "mam", "jja", "son")my_files_2 <- sample(file_all, 100000, replace = TRUE)microbenchmark(sotos = my_files_2[order(match(gsub(".*-(.*)-tas.*", "\", my_files_2), my_order))],sotos_2 = { split_files <- strsplit(my_files_2, "-")i3 <- unlist(lapply(split_files, "[[", 3))i4 <- unlist(lapply(split_files, "[[", 2))my_files_2[order(i4, match(i3, my_order))] },sotos_3 = { split_files <- strsplit(my_files_2, "-", fixed = TRUE)i3 <- unlist(lapply(split_files, "[[", 3))i4 <- unlist(lapply(split_files, "[[", 2))my_files_2[order(i4, match(i3, my_order))] },alt = my_files_2[unlist(lapply(my_order, \(n) which(grepl(n, my_files_2))))],tidyverse = my_files_2 %>%data.frame(model = .) %>%separate_wider_delim(model, delim = "-", names_sep = "", cols_remove = FALSE) %>%mutate(model3 = match(model3, my_order)) %>%arrange(model2, model3) %>%pull(modelmodel))
The result surprised me quite a bit
#> Unit: milliseconds#> expr min lq mean median uq max neval#> sotos 126.5779 127.5660 129.03582 128.17515 128.83190 199.8997 100#> sotos_2 403.7318 412.5335 430.80871 419.50455 433.35235 511.1252 100#> sotos_3 220.1570 228.3046 244.84893 235.50230 246.01510 314.8171 100#> alt 113.9459 114.7081 115.86338 115.50140 116.37555 121.4351 100#> tidyverse 75.5209 83.2439 93.82626 85.46735 92.48615 183.7602 100
Here are a few benchmark results based on the size of my_list_2:
my_files_2 <- sample(file_all, 100, replace = TRUE)#> Unit: microseconds#> expr min lq mean median uq max neval#> sotos 155.6 162.95 175.719 173.70 184.05 263.3 100#> sotos_2 399.6 416.35 446.641 427.35 448.80 705.4 100#> sotos_3 217.6 229.60 278.047 237.20 252.25 3846.8 100#> alt 139.6 150.00 169.377 154.75 160.30 1452.0 100#> tidyverse 7912.6 8265.80 8927.205 8529.85 8987.45 14056.3 100my_files_2 <- sample(file_all, 1000, replace = TRUE)#> Unit: milliseconds#> expr min lq mean median uq max neval#> sotos 1.2786 1.31485 1.368087 1.33890 1.37390 1.9579 100#> sotos_2 3.6715 3.77725 3.849989 3.83145 3.90755 4.3005 100#> sotos_3 1.8969 1.94865 2.004843 1.97525 2.01915 2.4871 100#> alt 1.1500 1.18695 1.229627 1.20130 1.22640 2.4962 100#> tidyverse 8.6329 8.99985 9.710873 9.21335 9.63085 14.6684 100my_files_2 <- sample(file_all, 10000, replace = TRUE)#> Unit: milliseconds#> expr min lq mean median uq max neval#> sotos 12.5224 12.81450 12.92100 12.89505 13.00510 13.5006 100#> sotos_2 37.5119 38.09420 38.98338 38.34025 38.76710 51.7555 100#> sotos_3 19.3073 19.70905 20.63162 19.88925 20.29015 25.1160 100#> alt 11.2854 11.54560 11.74392 11.63175 11.72850 15.7317 100#> tidyverse 14.4663 15.03035 16.42199 15.27875 15.49390 83.9643 100
At a sample size of ~20000 the tidyverse solution seems to take the lead.
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