对 R 中分成 N 个块的数据运行迭代回归 [英] Running iterated regressions for data divided into N chunks in R

问题描述

我有一个结构如下的数据框:

I have a dataframe structured like the following:

birthwt  tobacco01  pscore  pscoreblocks
3425     0          0.18    (0.177, 0.187]
3527     1          0.15    (0.158, 0.168]
1638     1          0.34    (0.335, 0.345]

birthwt 列是以克为单位测量出生体重的连续变量.烟草 01 列包含 0 或 1 的值. pscore 列包含 0 和 1 之间的概率值. pscoreblocks 获取 pscore 列并将其分解为 100 个相同大小的块.

The birthwt column is a continuous variable measuring birth weight in grams. The tobacco01 column contains values of 0 or 1. The pscore column contains probability values between 0 and 1. The pscoreblocks takes the pscore column and breaks it down into 100 equally sized blocks.

我试图找到一种有效的方法来为 pscoreblocks 中的每个块执行以下操作.我已经包含了如果我在整个数据集上运行它而不分区成块的代码.

I am trying to find an efficient way to do the following for each of the blocks in pscoreblocks. I have included the code that would work if I was running this on the entire dataset without partitioning into blocks.

1- 运行回归.

one <- lm(birthwt ~ tobacco01, dfc)

2- 在回归中取烟草01 变量的系数值.

2- Take the value of the coefficient on the tobacco01 variable in the regression.

two <- summary(one)$coefficients[2,1]

3- 将该系数值乘以:[(该块中烟草的人数== 1)+(该块中烟草的人数烟草 == 0 在那个块中)]/(在那个块中的总人数块)

3- Multiply that coefficient value by: [(the number of people for whom tobacco == 1 in that block) + (the number of people for whom tobacco == 0 in that block)] / (the total number of people in that block)

two_5 <- ((sum(dfc$tobacco01 == 1)) + (sum(dfc$tobacco01 == 0)))/ sum(dfc$tobacco)

three <- two*two_5

4- 最后，我希望能够将 (3) 中所有 100 个块的所有值相加.

4- Finally, I would like to be able to add up all the values from (3) for all 100 blocks.

我知道如何单独执行这些步骤中的每一个，但我不知道如何在 100 个单独的块上迭代它们.我尝试使用 group_by(pscoreblocks) 然后运行回归，但看起来 group_by() 和 lm() 不能很好地协同工作.我还考虑使用 pivot_longer() 为每个块创建一个单独的列，然后尝试使用该格式的数据运行回归.我非常感谢有关如何迭代所有 100 个块的任何建议.

I know how to do each of these steps individually, but I don't know how to iterate them over 100 separate blocks. I tried using group_by(pscoreblocks) and then running a regression, but it looks like group_by() and lm() do not work well together. I have also considered using pivot_longer() to create a separate column for each block and then trying to run the regressions with the data in that format. I'd really appreciate any suggestions for how to iterate over all 100 blocks.

数据:

> small <- dput(dfcsmall[1:40,])
structure(list(dbrwt = c(3629, 3005, 3459, 4520, 3095.17811313023, 
3714, 3515, 3232, 3686, 4281, 2645.29691556227, 3714, 3232, 3374, 
3856, 3997, 3515, 3714, 3459, 3232, 3884, 3235, 3008.94507753983, 
3799, 2940, 3389.51332290472, 3090, 1701, 3363, 3033, 2325, 3941, 
3657, 3600, 3005, 4054, 3856, 3402, 2694.09822203382, 3413.03869100037
), tobacco01 = c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 1, 1), pscore = c(0.00988756408875347, 0.183983728674846, 
0.24538311074894, 0.170701594663405, 0.179337494008595,         0.0770304781540708, 
0.164003166666384, 0.0773042518100593, 0.0804603038634144,     0.0611822720382283, 
0.481204657069376, 0.166016137665693, 0.107882394783232,     0.149799473798458, 
0.04130366288307, 0.0360272679038012, 0.476513676221723, 0.214910849480014, 
0.0687582392973688, 0.317662260996216, 0.206183065905609,     0.336553699970873, 
0.0559863953956171, 0.103064791185442, 0.0445362319933672,     0.17097032928289, 
0.245898950803051, 0.146235179401833, 0.284345485401689,     0.152121397241563, 
0.0395696572471225, 0.116669642645446, 0.0672219220193578,     0.297173652687617, 
0.436771917147971, 0.0517299620576624, 0.140760280612358,     0.179726730598874, 
0.0118610298424373, 0.162996197785343), pscoreblocks = structure(c(1L, 
19L, 25L, 18L, 19L, 8L, 17L, 8L, 9L, 7L, 49L, 17L, 11L, 16L, 
5L, 4L, 49L, 22L, 7L, 33L, 21L, 35L, 6L, 11L, 5L, 18L, 25L, 15L, 
29L, 16L, 5L, 12L, 7L, 31L, 45L, 6L, 15L, 19L, 2L, 17L), .Label = c("    [3.88e-05,0.0099]", 
"(0.0099,0.0198]", "(0.0198,0.0296]", "(0.0296,0.0395]", "    (0.0395,0.0493]", 
"(0.0493,0.0592]", "(0.0592,0.069]", "(0.069,0.0789]", "(0.0789,0.0888]", 
"(0.0888,0.0986]", "(0.0986,0.108]", "(0.108,0.118]", "(0.118,0.128]", 
"(0.128,0.138]", "(0.138,0.148]", "(0.148,0.158]", "(0.158,0.168]", 
"(0.168,0.177]", "(0.177,0.187]", "(0.187,0.197]", "(0.197,0.207]", 
"(0.207,0.217]", "(0.217,0.227]", "(0.227,0.237]", "(0.237,0.246]", 
"(0.246,0.256]", "(0.256,0.266]", "(0.266,0.276]", "(0.276,0.286]", 
"(0.286,0.296]", "(0.296,0.306]", "(0.306,0.315]", "(0.315,0.325]", 
"(0.325,0.335]", "(0.335,0.345]", "(0.345,0.355]", "(0.355,0.365]", 
"(0.365,0.375]", "(0.375,0.384]", "(0.384,0.394]", "(0.394,0.404]", 
"(0.404,0.414]", "(0.414,0.424]", "(0.424,0.434]", "(0.434,0.444]", 
"(0.444,0.453]", "(0.453,0.463]", "(0.463,0.473]", "(0.473,0.483]", 
"(0.483,0.493]", "(0.493,0.503]", "(0.503,0.513]", "(0.513,0.522]", 
"(0.522,0.532]", "(0.532,0.542]", "(0.542,0.552]", "(0.552,0.562]", 
"(0.562,0.572]", "(0.572,0.582]", "(0.582,0.591]", "(0.591,0.601]", 
"(0.601,0.611]", "(0.611,0.621]", "(0.621,0.631]", "(0.631,0.641]", 
"(0.641,0.651]", "(0.651,0.66]", "(0.66,0.67]", "(0.67,0.68]", 
"(0.68,0.69]", "(0.69,0.7]", "(0.7,0.71]", "(0.71,0.72]", "(0.72,0.73]", 
"(0.73,0.739]", "(0.739,0.749]", "(0.749,0.759]", "(0.759,0.769]", 
"(0.769,0.779]", "(0.779,0.789]", "(0.789,0.799]", "(0.799,0.808]", 
"(0.808,0.818]", "(0.818,0.828]", "(0.828,0.838]", "(0.838,0.848]", 
"(0.848,0.858]", "(0.858,0.868]", "(0.868,0.877]", "(0.877,0.887]", 
"(0.887,0.897]", "(0.897,0.907]", "(0.907,0.917]", "(0.917,0.927]", 
"(0.927,0.937]", "(0.937,0.946]", "(0.946,0.956]", "(0.956,0.966]", 
"(0.966,0.976]", "(0.976,0.986]"), class = "factor"), blocknumber = c(1L, 
19L, 25L, 18L, 19L, 8L, 17L, 8L, 9L, 7L, 49L, 17L, 11L, 16L, 
5L, 4L, 49L, 22L, 7L, 33L, 21L, 35L, 6L, 11L, 5L, 18L, 25L, 15L, 
29L, 16L, 5L, 12L, 7L, 31L, 45L, 6L, 15L, 19L, 2L, 17L)), row.names =     c(NA, 
-40L), class = c("tbl_df", "tbl", "data.frame"))

推荐答案

您可以创建一个函数来应用到每个 pscoreblocks.

You could create a function to apply to each pscoreblocks.

apply_model <- function(data) {
   one <- lm(birthwt ~ tobacco01, data)
   two <- summary(one)$coefficients[2,1]
   two_5 <- ((sum(data$tobacco01 == 1)) + (sum(data$tobacco01 == 0)))/ sum(data$tobacco)
   three <- two*two_5
   return(three)
}

将数据拆分为单独的数据帧并将此函数应用于每个块.

Split the data into spearate dataframe and apply this function to each chunk.

library(dplyr)
library(purrr)

dfc %>% group_split(pscoreblocks) %>% map(apply_model)
#OR
#dfc %>% group_split(pscoreblocks) %>% map_dbl(apply_model)

你也可以使用基础 R :

You can also use base R :

lapply(split(dfc, dfc$pscoreblocks), apply_model)

或者用 by :

by(dfc, dfc$pscoreblocks, apply_model)

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