How can I calculate the slope of multiple subsets of a data frame more efficiently? [duplicate]

问题

This question already has answers here:

Linear Regression and group by in R (10 answers)

Closed 4 years ago.

I have a data set that contains the optical absorption (a) across a range of wavelengths (wl) for 16 different samples (bod) on 5 different days (day). The dput output for samples (bod) 1 - 3 is below.

I need the slope of the natural log of the absorption by wavelength for each sample on each day.

My current approach has been to do this manually:

# calculate and extract the slope for each sample and date
s275.1.0 <- lm(log(a) ~ wl, data = spec275, subset = bod == 1 & day == "2014-06-10")
s275.1.0.slope <- coef(s275.1.0)["wl"]
s275.2.0 <- lm(log(a) ~ wl, data = spec275, subset = bod == 2 & day == "2014-06-10")
s275.2.0.slope <- coef(s275.1.0)["wl"]
# etc...
# combine slopes into a vector
s275.slopes <- c(s275.1.0, s275.2.0) # etc...

Obviously this is rather tedious. Is there a straightforward way to streamline this code to get R to iterate these calculations over all of the samples and days?

Dput Output

structure(list(bod = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), 
CPOM = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L), .Label = c("no", "yes"), class = "factor"), 
Nutrient = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L), .Label = c("no", "yes"), class = "factor"), 
day = structure(c(5L, 5L, 5L, 5L, 5L, 5L, 2L, 2L, 2L, 5L, 
2L, 2L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 
4L, 4L, 4L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 4L, 
4L, 4L, 4L, 4L, 4L, 4L, 1L, 1L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 5L, 5L, 
5L, 2L, 3L, 5L, 2L, 5L, 5L, 5L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 
4L, 4L, 4L, 4L, 4L), .Label = c("2014-06-10", "2014-06-12", 
"2014-06-17", "2014-06-24", "2014-07-01"), class = "factor"), 
wl = c(295L, 279L, 283L, 286L, 289L, 292L, 284L, 279L, 282L, 
276L, 288L, 291L, 294L, 276L, 275L, 278L, 281L, 284L, 287L, 
290L, 293L, 275L, 278L, 281L, 284L, 287L, 290L, 293L, 288L, 
281L, 284L, 287L, 277L, 290L, 294L, 282L, 283L, 287L, 289L, 
285L, 277L, 275L, 281L, 284L, 287L, 290L, 276L, 279L, 282L, 
285L, 288L, 291L, 294L, 293L, 275L, 278L, 281L, 284L, 287L, 
290L, 293L, 275L, 278L, 292L, 295L, 288L, 277L, 279L, 282L, 
285L, 277L, 280L, 283L, 286L, 289L, 292L, 295L, 284L, 287L, 
290L, 293L, 279L, 293L, 275L, 278L, 281L, 277L, 291L, 293L, 
276L, 286L, 289L, 295L, 279L, 287L, 275L, 281L, 284L, 292L, 
278L, 287L, 280L, 284L, 288L, 276L), abs = c(0.083, 0.105, 
0.098, 0.093, 0.089, 0.086, 0.206, 0.221, 0.212, 0.109, 0.196, 
0.19, 0.181, 0.436, 0.231, 0.224, 0.415, 0.397, 0.379, 0.361, 
0.341, 0.124, 0.12, 0.114, 0.109, 0.105, 0.1, 0.096, 0.111, 
0.124, 0.116, 0.113, 0.131, 0.108, 0.1, 0.123, 0.122, 0.115, 
0.112, 0.118, 0.131, 0.136, 0.106, 0.1, 0.096, 0.093, 0.131, 
0.128, 0.122, 0.117, 0.114, 0.11, 0.106, 0.086, 0.122, 0.118, 
0.113, 0.109, 0.105, 0.103, 0.099, 0.114, 0.111, 0.099, 0.095, 
0.103, 0.119, 0.116, 0.111, 0.107, 0.056, 0.055, 0.051, 0.048, 
0.046, 0.043, 0.043, 0.144, 0.139, 0.135, 0.13, 0.153, 0.099, 
0.097, 0.093, 0.088, 0.157, 0.102, 0.076, 0.159, 0.082, 0.078, 
0.074, 0.119, 0.108, 0.122, 0.116, 0.11, 0.08, 0.097, 0.086, 
0.094, 0.089, 0.084, 0.1), a = c(19.1149, 24.1815, 22.5694, 
21.4179, 20.4967, 19.8058, 47.4418, 50.8963, 48.8236, 25.1027, 
45.1388, 43.757, 41.6843, 100.4108, 53.1993, 51.5872, 95.5745, 
91.4291, 87.2837, 83.1383, 78.5323, 28.5572, 27.636, 26.2542, 
25.1027, 24.1815, 23.03, 22.1088, 25.5633, 28.5572, 26.7148, 
26.0239, 30.1693, 24.8724, 23.03, 28.3269, 28.0966, 26.4845, 
25.7936, 27.1754, 30.1693, 31.3208, 24.4118, 23.03, 22.1088, 
21.4179, 30.1693, 29.4784, 28.0966, 26.9451, 26.2542, 25.333, 
24.4118, 19.8058, 28.0966, 27.1754, 26.0239, 25.1027, 24.1815, 
23.7209, 22.7997, 26.2542, 25.5633, 22.7997, 21.8785, 23.7209, 
27.4057, 26.7148, 25.5633, 24.6421, 12.8968, 12.6665, 11.7453, 
11.0544, 10.5938, 9.9029, 9.9029, 33.1632, 32.0117, 31.0905, 
29.939, 35.2359, 22.7997, 22.3391, 21.4179, 20.2664, 36.1571, 
23.4906, 17.5028, 36.6177, 18.8846, 17.9634, 17.0422, 27.4057, 
24.8724, 28.0966, 26.7148, 25.333, 18.424, 22.3391, 19.8058, 
21.6482, 20.4967, 19.3452, 23.03)), .Names = c("bod", "CPOM", 
"Nutrient", "day", "wl", "abs", "a"), row.names = c(151L, 378L, 
382L, 385L, 388L, 391L, 406L, 522L, 525L, 618L, 774L, 777L, 780L, 
1105L, 1126L, 1129L, 1474L, 1477L, 1480L, 1483L, 1486L, 1656L, 
1659L, 1662L, 1665L, 1668L, 1671L, 1674L, 2717L, 2832L, 2835L, 
2838L, 2949L, 2963L, 3052L, 3143L, 3150L, 3169L, 3184L, 3194L, 
3213L, 3293L, 3724L, 3727L, 3730L, 3733L, 4582L, 4585L, 4588L, 
4591L, 4594L, 4597L, 4600L, 4829L, 5003L, 5006L, 5009L, 5012L, 
5015L, 5018L, 5021L, 5541L, 5544L, 5807L, 5810L, 6127L, 6185L, 
6200L, 6203L, 6206L, 6637L, 7006L, 7009L, 7012L, 7015L, 7018L, 
7021L, 7157L, 7524L, 7527L, 7530L, 7881L, 7884L, 7975L, 7978L, 
7981L, 8366L, 8369L, 8480L, 8486L, 8594L, 8597L, 8603L, 8844L, 
8852L, 8961L, 8967L, 8970L, 9158L, 9268L, 9304L, 9310L, 9314L, 
9345L, 9389L), class = "data.frame")

回答1:

This is a typical split-apply-combine type operation:

1. split the data into the relevant chunks, in this case by day then bod
2. apply a function to each chunk of data; in this case a linear model & extract the slope coefficient,
3. combine the slope parameters plus the aggregation data into a summary output object, usually a data frame.

There are many ways to do this, especially say using the plyr and more recently the dplyr packages, but I still feel a base R, by-hand version is instructive.

First, split the data up. Here I use split() and spl ends up being a list of data frames, one per combination of day and bod:

spl <- with(spec275, split(spec275, list(day = day, bod = bod)))

Now we want a wrapper function to fit the lm() model and return the slope coefficient. This wrapper does that:

coefLM <- function(x) {
  coef(lm(log(a) ~ wl, data = x))[2]
}

We apply, in this case via sapply(), but vapply() would be safer, the wrapper function to each component of spl.

coefs <- sapply(spl, coefLM)

We get back a vector of slopes:

head(coefs)

> head(coefs)
2014-06-10.1.wl 2014-06-12.1.wl 2014-06-17.1.wl 2014-06-24.1.wl 2014-07-01.1.wl 
    -0.01456420     -0.01285560     -0.01554979     -0.01446049     -0.01478953 
2014-06-10.2.wl 
    -0.01545840

It is then required to put this vector back into an object with the unique combinations of day and bod.

If you want all this together, do:

spl <- with(spec275, split(spec275, list(day = day, bod = bod)))
out <- unique(spec275[, c("day", "bod")])
out <- transform(out, slope = sapply(spl, coefLM))

> out
            day bod       slope
151  2014-07-01   1 -0.01456420
406  2014-06-12   1 -0.01285560
1105 2014-06-10   1 -0.01554979
1656 2014-06-24   1 -0.01446049
2717 2014-06-17   1 -0.01478953
3143 2014-06-12   2 -0.01545840
3724 2014-06-10   2 -0.01359000
4582 2014-06-17   2 -0.01197854
5003 2014-06-24   2 -0.01157193
5807 2014-07-01   2 -0.01237455
6637 2014-06-10   3 -0.01652268
7157 2014-06-12   3 -0.01177575
7884 2014-06-17   3 -0.01194247
7975 2014-07-01   3 -0.01388849
9158 2014-06-24   3 -0.01365501

回答2:

Suppose your dataset is named as "data":

library(dplyr)
library(broom) # to change model results into data frame
results = data %>% 
  group_by(bod, day) %>%
  do(tidy(lm(log(a) ~ wl, data = .))) %>% 
  filter(term == "wl") 

Results will be like this:

> head(results)
Source: local data frame [6 x 7]
Groups: bod, day

  bod    day     term    estimate    std.error statistic      p.value
1   1 2014-06-10   wl -0.01456420 0.0008484573 -17.16550 6.757132e-05
2   1 2014-06-12   wl -0.01285560 0.0002345532 -54.80887 2.476989e-09
3   1 2014-06-17   wl -0.01554979 0.0006418376 -24.22699 2.233117e-06
4   1 2014-06-24   wl -0.01446049 0.0002585413 -55.93107 3.455827e-08
5   1 2014-07-01   wl -0.01478953 0.0004929797 -30.00027 7.718303e-07
6   2 2014-06-10   wl -0.01545840 0.0007932582 -19.48723 6.567167e-06

回答3:

library(data.table)        
setDT(spec275)[,.(slope=coef(lm(log(a) ~ wl))["wl"]),by=c("bod","day")]
        bod        day       slope
     1:   1 2014-07-01 -0.01478953
     2:   1 2014-06-12 -0.01285560
     3:   1 2014-06-10 -0.01456420
     4:   1 2014-06-24 -0.01446049
     5:   1 2014-06-17 -0.01554979
     6:   2 2014-06-12 -0.01359000
     7:   2 2014-06-10 -0.01545840
     8:   2 2014-06-17 -0.01197854
     9:   2 2014-06-24 -0.01157193
    10:   2 2014-07-01 -0.01237455
    11:   3 2014-06-10 -0.01652268
    12:   3 2014-06-12 -0.01177575
    13:   3 2014-06-17 -0.01194247
    14:   3 2014-07-01 -0.01365501
    15:   3 2014-06-24 -0.01388849

baseR solution:

spec275$group<-with(spec275,interaction(bod,day))
do.call(rbind,lapply(unique(spec275$group),function(i){
   data.frame(cbind(model=paste0("model",i),slope=coef(lm(log(a) ~ wl, data = spec275[spec275$group==i,]))["wl"]))}))
                 model               slope
wl   model1.2014-07-01 -0.0147895254594997
wl1  model1.2014-06-12 -0.0128555970222881
wl2  model1.2014-06-10 -0.0145641963972831
wl3  model1.2014-06-24 -0.0144604920742437
wl4  model1.2014-06-17  -0.015549793252286
wl5  model2.2014-06-12 -0.0135899954428342
wl6  model2.2014-06-10 -0.0154584020879208
wl7  model2.2014-06-17 -0.0119785366650048
wl8  model2.2014-06-24 -0.0115719324383674
wl9  model2.2014-07-01 -0.0123745534390794
wl10 model3.2014-06-10 -0.0165226797671165
wl11 model3.2014-06-12 -0.0117757454366332
wl12 model3.2014-06-17 -0.0119424723122258
wl13 model3.2014-07-01  -0.013655010998267
wl14 model3.2014-06-24 -0.0138884860775378

回答4:

Using dplyr:

library(dplyr)
spec275 %>% group_by(bod, day) %>%
            do(model = lm(log(a) ~ wl, data = .)) %>%
            mutate(coef=coef(model)["wl"])

Gives:

head(results)
Source: local data frame [6 x 4]

  bod        day   model        coef
1   1 2014-06-10 <S3:lm> -0.01456420
2   1 2014-06-12 <S3:lm> -0.01285560
3   1 2014-06-17 <S3:lm> -0.01554979
4   1 2014-06-24 <S3:lm> -0.01446049
5   1 2014-07-01 <S3:lm> -0.01478953
6   2 2014-06-10 <S3:lm> -0.01545840

The model column contains the raw lm if you want other parameters.

来源：https://stackoverflow.com/questions/31059043/how-can-i-calculate-the-slope-of-multiple-subsets-of-a-data-frame-more-efficient