使用不同列计算值之间的差异，并使用R计算有间隙的值
使用不同列计算值之间的差异，并使用R计算有间隙的值
提问于 2019-03-01 10:12:04
有没有人能帮我弄清楚如何根据我的月度数据来计算差值？例如，我想按年计算每口井在1-7月、2-8月、3-9月等期间的地下水数值差异。注意:在某些年份中，将会缺少一些月份。任何tidyverse解决方案都将受到欢迎。
Well year month value
<dbl> <dbl> <fct> <dbl>
1 222 1995 February 8.53
2 222 1995 March 8.69
3 222 1995 April 8.92
4 222 1995 May 9.59
5 222 1995 June 9.59
6 222 1995 July 9.70
7 222 1995 August 9.66
8 222 1995 September 9.46
9 222 1995 October 9.49
10 222 1995 November 9.31
# ... with 18,400 more rows
df1 <- subset(df, month %in% c("February", "August"))
test <- df1 %>%
dcast(site + year + Well ~ month, value.var = "value") %>%
mutate(Diff = February - August)谢谢,
西蒙
回答 1
Stack Overflow用户
发布于 2019-03-01 11:12:17
因此，我尝试制作一个数据集，并使用dplyr创建一个解决方案。最佳实践是包含生成样本数据集的方法，因此请在以后的问题中这样做。
# load required library
library(dplyr)
# generate data set of all site, well, and month combinations
## define valid values
sites = letters[1:3]
wells = 1:5
months = month.name
## perform a series of merges
full_sites_wells_months_set <-
merge(sites, wells) %>%
dplyr::rename(sites = x, wells = y) %>% # this line and the prior could be replaced on your system with initial_tibble %>% dplyr::select(sites, wells) %>% unique()
merge(months) %>%
dplyr::rename(months = y) %>%
dplyr::arrange(sites, wells)
# create sample initial_tibble
## define fraction of records to simulate missing months
data_availability <- 0.8
initial_tibble <-
full_sites_wells_months_set %>%
dplyr::sample_frac(data_availability) %>%
dplyr::mutate(values = runif(nrow(full_sites_wells_months_set)*data_availability)) # generate random groundwater values
# generate final result by joining full expected set of sites, wells, and months to actual data, then group by sites and wells and perform lag subtraction
final_tibble <-
full_sites_wells_months_set %>%
dplyr::left_join(initial_tibble) %>%
dplyr::group_by(sites, wells) %>%
dplyr::mutate(trailing_difference_6_months = values - dplyr::lag(values, 6L))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/54936999
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