Question

我有一个名为/uploads/2018/1/1/pic1.jpg的df，每个法律都有一行（每个id一个）：

laws

我想从中创建第二个名为laws <- data.frame(id=c(1,2,3),beginyear=c(2001,2002,2005),endyear=c(2003,2005,2006), law1=c(0,0,1), law2=c(1,0,1))，每个id和年份都有一行：

idyear

我如何有效地编写一些代码来获取idyear <- data.frame(id=c(rep(1,6),rep(2,6),rep(3,6)), year=(rep(c(2001:2006),3)), law1=c(rep(0,16),1,1), law2=c(1,1,1,rep(0,13),1,1)) df的idyear df输出？如果laws是＆gt; = idyear$year并且laws$beginyear是＆lt; = idyear$year，则两个法律变量是指标变量== 1。

我是R的初学者，但我愿意尝试任何事情（申请，循环等）以使其发挥作用。

Answer 1

1）基础 expand.grid将创建所有id和year组合的18 x 2数据框，然后merge将合并它与laws一起回来。将law1和law2之间year和beginyear之间的任何endyear和beginyear条目清零。最后删除endyear和g <- with(laws, expand.grid(year = min(beginyear):max(endyear), id = id)) m <- merge(g, laws) m[m$year < m$beginyear | m$year > m$endyear, c("law1", "law2")] <- 0 m <- subset(m, select = - c(beginyear, endyear)) # check identical(m, idyear) ## [1] TRUE列。没有包使用。

library(magrittr)

laws %$%
     expand.grid(year = min(beginyear):max(endyear), id = id) %>%
     merge(laws) %$%
     { .[year < beginyear | year > endyear, c("law1", "law2")] <- 0; .} %>%
     subset(select = - c(beginyear, endyear))

2）magrittr 这与（1）的解决方案相同，只是我们使用magrittr管道来表达它。注意管道操作符的混合。

Memcached

更新：已修复。添加了（2）。

Answer 2

使用mapply功能可以提供帮助。

# Function to expand year between begin and end
gen_data <- function(x_id, x_beginyear, x_endyear, x_law1, x_law2){
  df <- data.frame(x_id, x_beginyear:x_endyear, x_law1, x_law2)
  df
}

idyearlst <- data.frame()

idyearlst <- rbind(idyearlst, mapply(gen_data, laws$id, laws$beginyear,
 laws$endyear, laws$law1, laws$law2))

# Finally convert list to data.frame
idyear <- setNames(do.call(rbind.data.frame, idyearlst), c("id", "year", "law1", "law2"))

Result will be like:
> idyear
     id year law1 law2
V1.1  1 2001    0    1
V1.2  1 2002    0    1
V1.3  1 2003    0    1
V2.4  2 2002    0    0
V2.5  2 2003    0    0
V2.6  2 2004    0    0
V2.7  2 2005    0    0
V3.8  3 2005    1    1
V3.9  3 2006    1    1

Answer 3

使用tidyverse的解决方案。最后一个as.data.frame()是可选的，只是将tbl转换为数据框。

library(tidyverse)

idyear <- laws %>%
  mutate(year = map2(beginyear, endyear, `:`)) %>%
  unnest() %>%
  complete(id, year = full_seq(year, period = 1L), fill = list(law1 = 0L, law2 = 0L)) %>%
  select(-beginyear, -endyear) %>%
  as.data.frame()
idyear
#    id year law1 law2
# 1   1 2001    0    1
# 2   1 2002    0    1
# 3   1 2003    0    1
# 4   1 2004    0    0
# 5   1 2005    0    0
# 6   1 2006    0    0
# 7   2 2001    0    0
# 8   2 2002    0    0
# 9   2 2003    0    0
# 10  2 2004    0    0
# 11  2 2005    0    0
# 12  2 2006    0    0
# 13  3 2001    0    0
# 14  3 2002    0    0
# 15  3 2003    0    0
# 16  3 2004    0    0
# 17  3 2005    1    1
# 18  3 2006    1    1

Answer 4

有点丑陋的方法，但我认为它会得到你所追求的，使用G. Grothendieck的g expand.grid数据框作为基础，以及你的laws数据框。 / p>

new.df <- data.frame(t(apply(g, 1, function(x){
  yearspan = laws[laws$id == x['id'], 'beginyear']:laws[laws$id == x['id'], 'endyear']
  law1 = laws$law1[laws$id == x['id'] & x['year'] %in% yearspan]
  law2 = laws$law2[laws$id == x['id'] & x['year'] %in% yearspan]
  x['law1'] = ifelse(length(law1 > 0), law1, 0)
  x['law2'] = ifelse(length(law2 > 0), law2, 0)
  return(x)
})))

> new.df
   id year law1 law2
1   1 2001    0    1
2   1 2002    0    1
3   1 2003    0    1
4   1 2004    0    0
5   1 2005    0    0
6   1 2006    0    0
7   2 2001    0    0
8   2 2002    0    0
9   2 2003    0    0
10  2 2004    0    0
11  2 2005    0    0
12  2 2006    0    0
13  3 2001    0    0
14  3 2002    0    0
15  3 2003    0    0
16  3 2004    0    0
17  3 2005    1    1
18  3 2006    1    1

图书馆：

dplyr（对于arrange，并非真的有必要）

数据：

laws <- data.frame(id=c(1,2,3),
                   beginyear=c(2001,2002,2005),
                   endyear=c(2003,2005,2006), 
                   law1=c(0,0,1), law2=c(1,0,1))

g <- with(laws, expand.grid(id = id, year = min(beginyear):max(endyear)))
g <- arrange(g, id)

将每个id和law（带有开始和结束年份）的行的数据框转换为每个id和year有一行的文件

4 个答案: