英文:
Rolling mean per group in tidyverse
问题
以下是您要翻译的内容：
I aggregate data per group and calculate means per group to ease visualization. Unfortunately, some of my groups are very large, some are rather empty. I like to have a rolling mean calculation to smooth the picture further. Here is similar data:
load package
library(haven)
read dta file from github
soep <- read_dta("https://github.com/MarcoKuehne/marcokuehne.github.io/blob/main/data/SOEP/soep_lebensz_en/soep_lebensz_en.dta?raw=true")
soep %>%
group_by(education, sex) %>%
summarise(across(satisf_org, mean, na.rm = TRUE),
n = n()) %>%
ggplot(aes(x = education, y = satisf_org, col = as.factor(sex))) +
geom_point() +
labs(title = "Mean Satisfaction per Education Level by Gender",
x = "Education", y = "Mean Satisfaction", color = "Gender")
The mean satisfaction at education 8.5 for females looks like an outlier. At every year of education, I assume that people are not too different to be summarized, i.e. calculate the mean satisfaction of all people at education 7, 8.5 and 9 (grouped by sex) and store it as rolling mean at 8.5 (grouped by sex).
Starting from standard grouped means:
soep %>%
group_by(education, sex) %>%
summarise(across(satisf_org, mean, na.rm = TRUE),
n = n())
A tibble: 28 × 4
Groups: education [14]
education sex satisf_org n
<dbl> <dbl+lbl> <dbl> <int>
1 7 0 [male] 6.16 73
2 7 1 [female] 6.59 113
3 8.5 0 [male] 7.16 37
4 8.5 1 [female] 8.56 18
5 9 0 [male] 6.88 430
6 9 1 [female] 7.00 633
7 10 0 [male] 7.19 144
8 10 1 [female] 7.36 221
9 10.5 0 [male] 6.96 1538
10 10.5 1 [female] 7.02 1493
… with 18 more rows
ℹ Use print(n = ...) to see more rows
Here are the numbers that I expect
soep %>%
filter(sex == 1) %>% # only looks at females
filter(education %in% c(7, 8.5, 9)) %>% # take education level before and after
summarise(mean(satisf_org)) # calculate the "rolling mean" per group
A tibble: 1 × 1
mean(satisf_org)
<dbl>
1 6.97
This is the kind of rolling mean per group that I expect per value, i.e. 6.97 instead of 8.56.
PS: In my real data, I investigate age in years and I usually have at least some people at all ages. So the rolling window can be -1 to +1 (numeric) instead of lead / lag neighbors.
英文:
I aggregate data per group and calculate means per group to ease visualization. Unfortunately, some of my groups are very large, some are rather empty. I like to have a rolling mean calculation to smooth the picture further. Here is similar data:
# load packagelibrary(haven)# read dta file from githubsoep <- read_dta("https://github.com/MarcoKuehne/marcokuehne.github.io/blob/main/data/SOEP/soep_lebensz_en/soep_lebensz_en.dta?raw=true")soep %>%group_by(education, sex) %>%summarise(across(satisf_org, mean, na.rm = TRUE),n = n()) %>%ggplot(aes(x = education, y = satisf_org, col = as.factor(sex))) +geom_point() +labs(title = "Mean Satisfaction per Education Level by Gender",x = "Education", y = "Mean Satisfaction", color = "Gender")
The mean satisfaction at education 8.5 for females looks like an outlier. At every year of education, I assume that people are not too different to be summarized, i.e. calculate the mean satisfaction of all people at education 7, 8.5 and 9 (grouped by sex) and store it as rolling mean at 8.5 (grouped by sex).
Starting from standard grouped means:
soep %>%group_by(education, sex) %>%summarise(across(satisf_org, mean, na.rm = TRUE),n = n())# A tibble: 28 × 4# Groups: education [14]education sex satisf_org n<dbl> <dbl+lbl> <dbl> <int>1 7 0 [male] 6.16 732 7 1 [female] 6.59 1133 8.5 0 [male] 7.16 374 8.5 1 [female] 8.56 185 9 0 [male] 6.88 4306 9 1 [female] 7.00 6337 10 0 [male] 7.19 1448 10 1 [female] 7.36 2219 10.5 0 [male] 6.96 153810 10.5 1 [female] 7.02 1493# … with 18 more rows# ℹ Use `print(n = ...)` to see more rows
Here are the numbers that I expect
soep %>%filter(sex == 1) %>% # only looks at femalesfilter(education %in% c(7, 8.5, 9)) %>% # take education level before and aftersummarise(mean(satisf_org)) # calculate the "rolling mean" per group# A tibble: 1 × 1`mean(satisf_org)`<dbl>1 6.97
This is the kind of rolling mean per group that I expect per value, i.e. 6.97 instead of 8.56.
PS: In my real data, I investigate age in years and I usually have at least some people at all ages. So the rolling window can be -1 to +1 (numeric) instead of lead / lag neighbours.
答案1
得分: 2
你可以按性别进行group_by操作，然后进行滚动平均计算：
library(dplyr)library(slider)soep %>%group_by(education, sex) %>%summarise(across(satisf_org, mean, na.rm = TRUE),n = n()) %>%group_by(sex) %>%mutate(rolling_mean = slide_dbl(satisf_org, mean, .before = 1, .after = 1))
输出：
# A tibble: 28 × 5# Groups: sex [2]education sex satisf_org n rolling_mean<dbl> <dbl+lbl> <dbl> <int> <dbl>1 7 0 [male] 6.16 73 6.662 7 1 [female] 6.59 113 7.573 8.5 0 [male] 7.16 37 6.734 8.5 1 [female] 8.56 18 7.385 9 0 [male] 6.88 430 7.086 9 1 [female] 7.00 633 7.647 10 0 [male] 7.19 144 7.018 10 1 [female] 7.36 221 7.139 10.5 0 [male] 6.96 1538 7.1410 10.5 1 [female] 7.02 1493 7.20# … with 18 more rows# ℹ Use `print(n = ...)` to see more rows
注意：这只是代码的翻译部分，不包括问题中的其他内容。
英文:
You can group_by sex and do a rolling average there:
library(dplyr)library(slider)soep %>%group_by(education, sex) %>%summarise(across(satisf_org, mean, na.rm = TRUE),n = n()) %>%group_by(sex) %>%mutate(rolling_mean = slide_dbl(satisf_org, mean, .before = 1, .after = 1))
output
# A tibble: 28 × 5# Groups: sex [2]education sex satisf_org n rolling_mean<dbl> <dbl+lbl> <dbl> <int> <dbl>1 7 0 [male] 6.16 73 6.662 7 1 [female] 6.59 113 7.573 8.5 0 [male] 7.16 37 6.734 8.5 1 [female] 8.56 18 7.385 9 0 [male] 6.88 430 7.086 9 1 [female] 7.00 633 7.647 10 0 [male] 7.19 144 7.018 10 1 [female] 7.36 221 7.139 10.5 0 [male] 6.96 1538 7.1410 10.5 1 [female] 7.02 1493 7.20# … with 18 more rows# ℹ Use `print(n = ...)` to see more rows
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