我有一个大型数据集，需要分成多个平衡集。
该集合如下所示：
> data<-matrix(runif(4000, min=0, max=10), nrow=500, ncol=8 )
> colnames(data)<-c("A","B","C","D","E","F","G","H")
每个包含例如 20 行的集合需要在多个变量之间进行平衡，以便每个子集最终具有与所有其他子集相比包含在其子组中的 B、C、D 的相似平均值。
有没有办法用 R 做到这一点？任何建议将不胜感激。先感谢您！
2 回答
0
library(tidyverse)
# Reproducible data
set.seed(2)
data<-matrix(runif(4000, min=0, max=10), nrow=500, ncol=8 )
colnames(data)<-c("A","B","C","D","E","F","G","H")
data=as.data.frame(data)
更新的答案
如果您想将给定行的观察结果放在一起，则可能无法在每列内的集合中获得相似的平均值。有 8 列（如在您的示例数据中），您需要 25 个 20 行集，其中每列 A 集具有相同的均值，每列 B 集具有相同的均值，等等。这是很多约束。然而，可能有一些算法可以找到最小化集合均值差异的集合成员分配时间表。
但是，如果您可以从每列中分别获取 20 个观察值，而不考虑它来自哪一行，那么这里有一个选项：
# Group into sets with same means
same_means = data %>%
gather(key, value) %>%
arrange(value) %>%
group_by(key) %>%
mutate(set = c(rep(1:25, 10), rep(25:1, 10)))
# Check means by set for each column
same_means %>%
group_by(key, set) %>%
summarise(mean=mean(value)) %>%
spread(key, mean) %>% as.data.frame
set A B C D E F G H 1 1 4.940018 5.018584 5.117592 4.931069 5.016401 5.171896 4.886093 5.047926 2 2 4.946496 5.018578 5.124084 4.936461 5.017041 5.172817 4.887383 5.048850 3 3 4.947443 5.021511 5.125649 4.929010 5.015181 5.173983 4.880492 5.044192 4 4 4.948340 5.014958 5.126480 4.922940 5.007478 5.175898 4.878876 5.042789 5 5 4.943010 5.018506 5.123188 4.924283 5.019847 5.174981 4.869466 5.046532 6 6 4.942808 5.019945 5.123633 4.924036 5.019279 5.186053 4.870271 5.044757 7 7 4.945312 5.022991 5.120904 4.919835 5.019173 5.187910 4.869666 5.041317 8 8 4.947457 5.024992 5.125821 4.915033 5.016782 5.187996 4.867533 5.043262 9 9 4.936680 5.020040 5.128815 4.917770 5.022527 5.180950 4.864416 5.043587 10 10 4.943435 5.022840 5.122607 4.921102 5.018274 5.183719 4.872688 5.036263 11 11 4.942015 5.024077 5.121594 4.921965 5.015766 5.185075 4.880304 5.045362 12 12 4.944416 5.024906 5.119663 4.925396 5.023136 5.183449 4.887840 5.044733 13 13 4.946751 5.020960 5.127302 4.923513 5.014100 5.186527 4.889140 5.048425 14 14 4.949517 5.011549 5.127794 4.925720 5.006624 5.188227 4.882128 5.055608 15 15 4.943008 5.013135 5.130486 4.930377 5.002825 5.194421 4.884593 5.051968 16 16 4.939554 5.021875 5.129392 4.930384 5.005527 5.197746 4.883358 5.052474 17 17 4.935909 5.019139 5.131258 4.922536 5.003273 5.204442 4.884018 5.059162 18 18 4.935830 5.022633 5.129389 4.927106 5.008391 5.210277 4.877859 5.054829 19 19 4.936171 5.025452 5.127276 4.927904 5.007995 5.206972 4.873620 5.054192 20 20 4.942925 5.018719 5.127394 4.929643 5.005699 5.202787 4.869454 5.055665 21 21 4.941351 5.014454 5.125727 4.932884 5.008633 5.205170 4.870352 5.047728 22 22 4.933846 5.019311 5.130156 4.923804 5.012874 5.213346 4.874263 5.056290 23 23 4.928815 5.021575 5.139077 4.923665 5.017180 5.211699 4.876333 5.056836 24 24 4.928739 5.024419 5.140386 4.925559 5.012995 5.214019 4.880025 5.055182 25 25 4.929357 5.025198 5.134391 4.930061 5.008571 5.217005 4.885442 5.062630
原始答案
# Randomly group data into 20-row groups
set.seed(104)
data = data %>%
mutate(set = sample(rep(1:(500/20), each=20)))
head(data)
A B C D E F G H set 1 1.848823 6.920055 3.2283369 6.633721 6.794640 2.0288792 1.984295 2.09812642 10 2 7.023740 5.599569 0.4468325 5.198884 6.572196 0.9269249 9.700118 4.58840437 20 3 5.733263 3.426912 7.3168797 3.317611 8.301268 1.4466065 5.280740 0.09172101 19 4 1.680519 2.344975 4.9242313 6.163171 4.651894 2.2253335 1.175535 2.51299726 25 5 9.438393 4.296028 2.3563249 5.814513 1.717668 0.8130327 9.430833 0.68269106 19 6 9.434750 7.367007 1.2603451 5.952936 3.337172 5.2892300 5.139007 6.52763327 5
# Mean by set for each column
data %>% group_by(set) %>%
summarise_all(mean)
set A B C D E F G H 1 1 5.240236 6.143941 4.638874 5.367626 4.982008 4.200123 5.521844 5.083868 2 2 5.520983 5.257147 5.209941 4.504766 4.231175 3.642897 5.578811 6.439491 3 3 5.943011 3.556500 5.366094 4.583440 4.932206 4.725007 5.579103 5.420547 4 4 4.729387 4.755320 5.582982 4.763171 5.217154 5.224971 4.972047 3.892672 5 5 4.824812 4.527623 5.055745 4.556010 4.816255 4.426381 3.520427 6.398151 6 6 4.957994 7.517130 6.727288 4.757732 4.575019 6.220071 5.219651 5.130648 7 7 5.344701 4.650095 5.736826 5.161822 5.208502 5.645190 4.266679 4.243660 8 8 4.003065 4.578335 5.797876 4.968013 5.130712 6.192811 4.282839 5.669198 9 9 4.766465 4.395451 5.485031 4.577186 5.366829 5.653012 4.550389 4.367806 10 10 4.695404 5.295599 5.123817 5.358232 5.439788 5.643931 5.127332 5.089670 # ... with 15 more rows
如果数据框中的总行数不能被每个集合中所需的行数整除，则可以在创建集合时执行以下操作：
data = data %>%
mutate(set = sample(rep(1:ceiling(500/20), each=20))[1:n()])
在这种情况下，集合大小会随着数据行数不能被每个集合中所需的行数整除而略有不同。
于 2017-10-05T16:55:40.120 回答
0
对于处于类似位置的人来说，以下方法可能值得尝试。
它基于groupdata2'sfold()函数中的数值平衡，允许为单个列创建具有平衡平均值的组。通过标准化每一列并在数值上平衡它们的行总和，我们可能会增加在各个列中获得平衡均值的机会。
我将这种方法与随机创建组进行了几次比较，并选择了均值方差最小的拆分。它似乎好一点，但我不太相信这将适用于所有情况。
# Attach dplyr and groupdata2
library(dplyr)
library(groupdata2)
set.seed(1)
# Create the dataset
data <- matrix(runif(4000, min = 0, max = 10), nrow = 500, ncol = 8)
colnames(data) <- c("A", "B", "C", "D", "E", "F", "G", "H")
data <- dplyr::as_tibble(data)
# Standardize all columns and calculate row sums
data_std <- data %>%
dplyr::mutate_all(.funs = function(x){(x-mean(x))/sd(x)}) %>%
dplyr::mutate(total = rowSums(across(where(is.numeric))))
# Create groups (new column called ".folds")
# We numerically balance the "total" column
data_std <- data_std %>%
groupdata2::fold(k = 25, num_col = "total") # k = 500/20=25
# Transfer the groups to the original (non-standardized) data frame
data$group <- data_std$.folds
# Check the means
data %>%
dplyr::group_by(group) %>%
dplyr::summarise_all(.funs = mean)
> # A tibble: 25 x 9
> group A B C D E F G H
> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
> 1 1 4.48 5.05 4.80 5.65 5.04 4.60 5.12 4.85
> 2 2 5.57 5.17 3.21 5.46 4.46 5.89 5.06 4.79
> 3 3 4.33 6.02 4.57 6.18 4.76 3.79 5.94 3.71
> 4 4 4.51 4.62 4.62 5.27 4.65 5.41 5.26 5.23
> 5 5 4.55 5.10 4.19 5.41 5.28 5.39 5.57 4.23
> 6 6 4.82 4.74 6.10 4.34 4.82 5.08 4.89 4.81
> 7 7 5.88 4.49 4.13 3.91 5.62 4.75 5.46 5.26
> 8 8 4.11 5.50 5.61 4.23 5.30 4.60 4.96 5.35
> 9 9 4.30 3.74 6.45 5.60 3.56 4.92 5.57 5.32
> 10 10 5.26 5.50 4.35 5.29 4.53 4.75 4.49 5.45
> # … with 15 more rows
# Check the standard deviations of the means
# Could be used to compare methods
data %>%
dplyr::group_by(group) %>%
dplyr::summarise_all(.funs = mean) %>%
dplyr::summarise(across(where(is.numeric), sd))
> # A tibble: 1 x 8
> A B C D E F G H
> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
> 1 0.496 0.546 0.764 0.669 0.591 0.611 0.690 0.475
不过，最好在标准化数据上比较不同方法的均值和均值方差（或上述标准差）。在这种情况下，可以计算方差的总和并将其最小化。
data_std %>%
dplyr::select(-total) %>%
dplyr::group_by(.folds) %>%
dplyr::summarise_all(.funs = mean) %>%
dplyr::summarise(across(where(is.numeric), sd)) %>%
sum()
> 1.643989
比较多个平衡拆分
该fold()功能允许一次创建多个独特的分组因素（拆分）。因此，在这里，我将执行 20 次数值平衡拆分，并找到均值标准差之和最小的分组。我将进一步将其转换为函数。
create_multi_balanced_groups <- function(data, cols, k, num_tries){
# Extract the variables of interest
# We assume these are numeric but we could add a check
data_to_balance <- data[, cols]
# Standardize all columns
# And calculate rowwise sums
data_std <- data_to_balance %>%
dplyr::mutate_all(.funs = function(x){(x-mean(x))/sd(x)}) %>%
dplyr::mutate(total = rowSums(across(where(is.numeric))))
# Create `num_tries` unique numerically balanced splits
data_std <- data_std %>%
groupdata2::fold(
k = k,
num_fold_cols = num_tries,
num_col = "total"
)
# The new fold column names ".folds_1", ".folds_2", etc.
fold_col_names <- paste0(".folds_", seq_len(num_tries))
# Remove total column
data_std <- data_std %>%
dplyr::select(-total)
# Calculate score for each split
# This could probably be done more efficiently without a for loop
variance_scores <- c()
for (fcol in fold_col_names){
score <- data_std %>%
dplyr::group_by(!!as.name(fcol)) %>%
dplyr::summarise(across(where(is.numeric), mean)) %>%
dplyr::summarise(across(where(is.numeric), sd)) %>%
sum()
variance_scores <- append(variance_scores, score)
}
# Get the fold column with the lowest score
lowest_fcol_index <- which.min(variance_scores)
best_fcol <- fold_col_names[[lowest_fcol_index]]
# Add the best fold column / grouping factor to the original data
data[["group"]] <- data_std[[best_fcol]]
# Return the original data and the score of the best fold column
list(data, min(variance_scores))
}
# Run with 20 splits
set.seed(1)
data_grouped_and_score <- create_multi_balanced_groups(
data = data,
cols = c("A", "B", "C", "D", "E", "F", "G", "H"),
k = 25,
num_tries = 20
)
# Check data
data_grouped_and_score[[1]]
> # A tibble: 500 x 9
> A B C D E F G H group
> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <fct>
> 1 5.86 6.54 0.500 2.88 5.70 9.67 2.29 3.01 2
> 2 0.0895 4.69 5.71 0.343 8.95 7.73 5.76 9.58 1
> 3 2.94 1.78 2.06 6.66 9.54 0.600 4.26 0.771 16
> 4 2.77 1.52 0.723 8.11 8.95 1.37 6.32 6.24 7
> 5 8.14 2.49 0.467 8.51 0.889 6.28 4.47 8.63 13
> 6 2.60 8.23 9.17 5.14 2.85 8.54 8.94 0.619 23
> 7 7.24 0.260 6.64 8.35 8.59 0.0862 1.73 8.10 5
> 8 9.06 1.11 6.01 5.35 2.01 9.37 7.47 1.01 1
> 9 9.49 5.48 3.64 1.94 3.24 2.49 3.63 5.52 7
> 10 0.731 0.230 5.29 8.43 5.40 8.50 3.46 1.23 10
> # … with 490 more rows
# Check score
data_grouped_and_score[[2]]
> 1.552656
通过注释掉该num_col = "total"行，我们可以在没有数值平衡的情况下运行它。对我来说，这给出了 1.615257 的分数。
免责声明：我是groupdata2包的作者。该fold()函数还可以平衡一个分类列cat_col（id_col还有一个非常相似的partition()功能。
于 2021-07-02T13:50:53.797 回答