英文:
Looking for a more efficient way of doing this for-loop in R
问题
我的数据集有超过300万条记录，所以这个循环花费了很长时间。
我想要创建一个bout_len_tracker变量，用于计算连续的行数，其中相同的主题（SUBJECT）具有VECTORMAGNITUDECOUNTS >= 1853。
sub_study$bout_len_tracker <- 0for (i in 2:nrow(sub_study)) {if ((sub_study$VECTORMAGNITUDECOUNTS[i] >= 1853) & (sub_study$SUBJECT[i] == sub_study$SUBJECT[i-1]))sub_study$bout_len_tracker[i] <- sub_study$bout_len_tracker[i-1] + 1}
英文:
My data set has over 3 million records so this loop is taking forever.
I want to create a bout_len_tracker variable that counts how many consecutive rows for the same SUBJECT have VECTORMAGNITUDECOUNTS >= 1853.
sub_study$bout_len_tracker <- 0for ( i in 2:nrow(sub_study) ) {if ( (sub_study$VECTORMAGNITUDECOUNTS[i] >= 1853) & (sub_study$SUBJECT[i] == sub_study$SUBJECT[i-1]) )sub_study$bout_len_tracker[i] <- sub_study$bout_len_tracker[i-1]+1}
答案1
得分: 1
你可以将 VECTORMAGNITUDECOUNTS >= 1853 的 cumsum 放入 ave 中。这里有一个示例：
dat$len <- with(dat, ave(VECTORMAGNITUDECOUNTS >= 1853, id, FUN=cumsum))dat# id t VECTORMAGNITUDECOUNTS len# 1 1 1 17 1# 2 1 2 18 2# 3 1 3 5 2# 4 2 1 5 0# 5 2 2 17 1# 6 2 3 14 2# 7 3 1 1 0# 8 3 2 15 1# 9 3 3 20 2# 10 4 1 10 1# 11 4 2 7 1# 12 4 3 18 2# 13 5 1 4 0# 14 5 2 4 0# 15 5 3 15 1
如果 n 中有 NA 值，
dat$n[sample.int(nrow(dat), nrow(dat)*.2)] <- NA
你可以将其扩展为：
dat$len <- with(dat, ave(VECTORMAGNITUDECOUNTS >= 1853, id, FUN=\(x) cumsum(replace(x, is.na(x), 0))))
数据:
dat <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,4L, 4L, 5L, 5L, 5L), t = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,1L, 2L, 3L, 1L, 2L, 3L), n = c(17L, 18L, 5L, 5L, 17L, 14L, 1L,15L, 20L, 10L, 7L, 18L, 4L, 4L, 15L)), out.attrs = list(dim = c(id = 5L,t = 3L), dimnames = list(id = c("id=1", "id=2", "id=3", "id=4","id=5"), t = c("t=1", "t=2", "t=3"))), row.names = c(NA, -15L), class = "data.frame")
英文:
You could put cumsum of VECTORMAGNITUDECOUNTS >= 1853 in ave. Here an example:
dat$len <- with(dat, ave(n >= 10, id, FUN=cumsum))dat# id t n len# 1 1 1 17 1# 2 1 2 18 2# 3 1 3 5 2# 4 2 1 5 0# 5 2 2 17 1# 6 2 3 14 2# 7 3 1 1 0# 8 3 2 15 1# 9 3 3 20 2# 10 4 1 10 1# 11 4 2 7 1# 12 4 3 18 2# 13 5 1 4 0# 14 5 2 4 0# 15 5 3 15 1
If there are NAs in n,
dat$n[sample.int(nrow(dat), nrow(dat)*.2)] <- NA
you can expand this to:
dat$len <- with(dat, ave(n >= 10, id, FUN=\(x) cumsum(replace(x, is.na(x), 0))))
Data:
dat <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L,4L, 4L, 5L, 5L, 5L), t = c(1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,1L, 2L, 3L, 1L, 2L, 3L), n = c(17L, 18L, 5L, 5L, 17L, 14L, 1L,15L, 20L, 10L, 7L, 18L, 4L, 4L, 15L)), out.attrs = list(dim = c(id = 5L,t = 3L), dimnames = list(id = c("id=1", "id=2", "id=3", "id=4","id=5"), t = c("t=1", "t=2", "t=3"))), row.names = c(NA, -15L), class = "data.frame")
答案2
得分: 1
以下是代码的翻译部分：
library(data.table)setDT(dat)[, tracker := (n >= 10) * rowid(n >= 10), id][]
如果需要进一步的帮助，请随时提问。
英文:
library(data.table)
setDT(dat)[,tracker:= (n>=10) * rowid(n >= 10),id][]id t n tracker1: 1 1 17 12: 1 2 18 23: 1 3 5 04: 2 1 5 05: 2 2 17 16: 2 3 14 27: 3 1 1 08: 3 2 15 19: 3 3 20 210: 4 1 10 111: 4 2 7 012: 4 3 18 213: 5 1 4 014: 5 2 4 015: 5 3 15 1
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