使用plyr计算大于95％的分位数时出现错误 [英] Error when calculating values greater than 95% quantile using plyr
问题描述
我的数据结构如下:
Individ <- data.frame(Participant = c("Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill",
"Harry", "Harry", "Harry", "Harry","Harry", "Harry", "Harry", "Harry", "Paul", "Paul", "Paul", "Paul"),
Time = c(1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4),
Condition = c("Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr",
"Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr"),
Power = c(400, 250, 180, 500, 300, 450, 600, 512, 300, 500, 450, 200, 402, 210, 130, 520, 310, 451, 608, 582, 390, 570, NA, NA))
使用dplyr我通过以下代码应用滚动平均值(从2到4秒):
Using dplyr I apply a rolling average (from 2 to 4 seconds) via the following code:
for (summaryFunction in c("mean")) {
for ( i in seq(2, 4, by = 1)) {
tempColumn <- Individ %>%
group_by(Participant) %>%
transmute(rollapply(Power,
width = i,
FUN = summaryFunction,
align = "right",
fill = NA,
na.rm = T))
colnames(tempColumn)[2] <- paste("Rolling", summaryFunction, as.character(i), sep = ".")
Individ <- bind_cols(Individ, tempColumn[2])
}
}
我现在希望计算每个滚动平均值中每个Participant的Power的前5％.为了计算这一点，我使用:
I now wish to calculate the top 5% of Power for each Participant across each of the rolling averages. To compute this, I use:
Output = ddply(Individ, .(Participant, Condition), summarise,
TwoSec <- Rolling.mean.2 > quantile(Rolling.mean.2 , 0.95, na.rm = TRUE))
但是，我最后得到一列TRUE或FALSE的列.相反，我追求的是前5％的实际值.我该怎么做呢?是否还有一种更简单的方法可以按参与者和条件遍历每个滚动平均值列，以找到每个平均值的前5％?
However, I end up with a column that states TRUE or FALSE. Instead, I am after the actual values that are in the top 5%. How do I do this? Is there also an easier way to iterate over each rolling mean column, by participant and condition, to find the top 5% in each?
谢谢！
推荐答案
最好获得滚动数据表，这使分位数的计算工作变得容易得多.
It's good that you got your rolling data table, that made the job of calculating the quantiles a lot easier.
第1步:按参加者，条件，位置分组
Individ <- data.frame(Participant = c("Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill", "Bill",
"Harry", "Harry", "Harry", "Harry","Harry", "Harry", "Harry", "Harry", "Paul", "Paul", "Paul", "Paul"),
Time = c(1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4),
Condition = c("Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr",
"Placebo", "Placebo", "Placebo", "Placebo", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr", "Expr"),
Location = c("Home", "Home", "Home", "Home", "Away", "Away", "Away", "Away", "Home", "Home", "Home", "Home",
"Home", "Home", "Home", "Home", "Away", "Away", "Away", "Away", "Home", "Home", "Home", "Home"),
Power = c(400, 250, 180, 500, 300, 450, 600, 512, 300, 500, 450, 200, 402, 210, 130, 520, 310, 451, 608, 582, 390, 570, NA, NA))
library(dplyr)
library(zoo)
for (summaryFunction in c("mean")) {
for ( i in seq(2, 4, by = 1)) {
tempColumn <- Individ %>%
group_by(Participant) %>%
transmute(rollapply(Power,
width = i,
FUN = summaryFunction,
align = "right",
fill = NA,
na.rm = T))
colnames(tempColumn)[2] <- paste("Rolling", summaryFunction, as.character(i), sep = ".")
Individ <- bind_cols(Individ, tempColumn[2])
}
}
Individ
Participant Time Condition Location Power Rolling.mean.2 Rolling.mean.3 Rolling.mean.4
(fctr) (dbl) (fctr) (fctr) (dbl) (dbl) (dbl) (dbl)
1 Bill 1 Placebo Home 400 NA NA NA
2 Bill 2 Placebo Home 250 325 NA NA
3 Bill 3 Placebo Home 180 215 276.6667 NA
4 Bill 4 Placebo Home 500 340 310.0000 332.5
5 Bill 1 Expr Away 300 400 326.6667 307.5
6 Bill 2 Expr Away 450 375 416.6667 357.5
7 Bill 3 Expr Away 600 525 450.0000 462.5
8 Bill 4 Expr Away 512 556 520.6667 465.5
9 Bill 1 Expr Home 300 406 470.6667 465.5
10 Bill 2 Expr Home 500 400 437.3333 478.0
在获得所有7或8列(此数据集包括位置)之后，它也回答了另一个问题，在新的Individ数据集中，这是我为解决您的问题所做的事情.我100％肯定有一种更干净，更有效的方法来执行此操作，但是这里有逻辑，应该可以输出.
After getting all 7 or 8 columns (this dataset includes Location), so it answers the other question as well, in the new Individ dataset, here's what I did to solve your problem. I'm 100% sure there is a cleaner and more efficient way to do this, but there is logic here and it should output fine.
第2步:获取组的分位数
library(plyr)
Individ[is.na(Individ)]<- 0
Top_percentiles <- ddply(Individ,
c("Participant", "Condition", "Location"),
summarise,
Power2 = quantile(Rolling.mean.2, .95),
Power3 = quantile(Rolling.mean.3, .95),
Power4 = quantile(Rolling.mean.4, .95)
)
Top_percentiles
Participant Condition Location Power2 Power3 Power4
1 Bill Expr Away 551.350 510.0667 465.050
2 Bill Expr Home 464.650 465.6667 476.125
3 Bill Placebo Home 337.750 305.0000 282.625
4 Harry Expr Away 585.175 533.4000 485.425
5 Harry Placebo Home 322.150 280.7667 268.175
6 Paul Expr Home 556.500 556.5000 408.000
这是每个组的前5％的阈值和相应的滚动平均值.
which is the threshold for the top 5% for each group and the corresponding rolling averages.
现在剩下要做的就是计算数据集中每个阈值以上的观测值.
Now the only thing left to do is calculate the observations in your dataset that are above each threshold.
第3步:将滚动平均值列与原始数据集匹配
像这样的事情我正在修补.
Something like this is kinda what I am tinkering around with.
Individ$Power2 <- Top_percentiles$Power2[match(Individ$Participant, Top_percentiles$Participant) &&
match(Individ$Condition, Top_percentiles$Condition) &&
match(Individ$Location, Top_percentiles$Location)]
Individ$Power3 <- Top_percentiles$Power3[match(Individ$Participant, Top_percentiles$Participant) &&
match(Individ$Condition, Top_percentiles$Condition) &&
match(Individ$Location, Top_percentiles$Location)]
Individ$Power4 <- Top_percentiles$Power4[match(Individ$Participant, Top_percentiles$Participant) &&
match(Individ$Condition, Top_percentiles$Condition) &&
match(Individ$Location, Top_percentiles$Location)]
Individ
Participant Time Condition Location Power Rolling.mean.2 Rolling.mean.3 Rolling.mean.4 Power2 Power3
(fctr) (dbl) (fctr) (fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 Bill 1 Placebo Home 400 0 0.0000 0.0 551.350 510.0667
2 Bill 2 Placebo Home 250 325 0.0000 0.0 464.650 465.6667
3 Bill 3 Placebo Home 180 215 276.6667 0.0 337.750 305.0000
4 Bill 4 Placebo Home 500 340 310.0000 332.5 585.175 533.4000
5 Bill 1 Expr Away 300 400 326.6667 307.5 322.150 280.7667
6 Bill 2 Expr Away 450 375 416.6667 357.5 556.500 556.5000
7 Bill 3 Expr Away 600 525 450.0000 462.5 551.350 510.0667
8 Bill 4 Expr Away 512 556 520.6667 465.5 464.650 465.6667
9 Bill 1 Expr Home 300 406 470.6667 465.5 337.750 305.0000
10 Bill 2 Expr Home 500 400 437.3333 478.0 585.175 533.4000
我在这里的想法是将分位数列匹配到Individ数据集上.
My idea here was to match the quantile columns onto the Individ dataset.
第4步:过滤数据集
这应该可以让您想要.
选项1:三个独立的数据集
top_percentile_2sec <- Individ %>% filter(Rolling.mean.2 >= Power2)
top_percentile_3sec <- Individ %>% filter(Rolling.mean.3 >= Power3)
top_percentile_4sec <- Individ %>% filter(Rolling.mean.4 >= Power4)
选项2:一个大的合并数据集
top_percentile_all_times <- Individ %>% filter(Rolling.mean.2 >= Power2 | Rolling.mean.3 >= Power3 | Rolling.mean.4 >= Power4)
top_percentile_all_times
Participant Time Condition Location Power Rolling.mean.2 Rolling.mean.3 Rolling.mean.4 Power2 Power3
(fctr) (dbl) (fctr) (fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 Bill 1 Expr Away 300 400.0 326.6667 307.50 322.15 280.7667
2 Bill 4 Expr Away 512 556.0 520.6667 465.50 464.65 465.6667
3 Bill 1 Expr Home 300 406.0 470.6667 465.50 337.75 305.0000
4 Bill 3 Expr Home 450 475.0 416.6667 440.50 322.15 280.7667
5 Harry 1 Expr Away 310 415.0 320.0000 292.50 322.15 280.7667
6 Harry 3 Expr Away 608 529.5 456.3333 472.25 551.35 510.0667
7 Harry 4 Expr Away 582 595.0 547.0000 487.75 464.65 465.6667
8 Paul 3 Expr Home 0 570.0 480.0000 0.00 322.15 280.7667
9 Paul 4 Expr Home 0 0.0 570.0000 480.00 556.50 556.5000
下面的链接极大地帮助了我.
Below is a link that greatly helped me out.
how to calculate 95th percentile of values with grouping variable in R or Excel
这也可以从其他帖子中解决您的问题吗?
这篇关于使用plyr计算大于95％的分位数时出现错误的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
