问题描述
数据框df有13个变量如下，
column 1-id
column 2-pr_1 （pr_1 to pr_12 are binary variables)
column 3-pr_2
column 4-pr_3
column 5-pr_4
column 6-pr_5
column 7-pr_6
column 8-pr_7
column 9-pr_8
column10-pr_9
column11-pr_10
column12-pr_11
column13-pr_12
Now,a variable "try" need to be created with
the following rule within the data frame and
for each observation,1)-The value of pr_1 always equals to 1.
2)-If all elements from pr_1 to pr_12 are 1,then try=13
3)-If there is a missing value(NA) between pr_1 to pr_12,then try= NA
4)-If the 1st 0 occurs right after the last 1,for example,the 1st 0 occurs in the variable pr_6 and the last 1 is in the variable pr_5,then the value of "try" should equal to 6 (6=5+1).
也就是说，“try”的值应该等于连续重复1次（重复次数中没有0或NA）加1。
带有新变量“try”的新数据集如下所示，
id pr_1 pr_2 pr_3 pr_4 pr_5 pr_6 pr_7 pr_8 pr_9 pr_10 pr_11 pr_12 try
j01 1 1 1 1 1 0 0 0 0 0 0 0 6
j02 1 1 1 0 0 0 0 0 0 0 0 0 4
j03 1 0 0 0 0 0 0 0 0 0 0 0 2
j04 1 1 1 1 1 1 1 1 1 1 1 1 13
j05 1 1 1 1 1 1 1 1 NA 1 1 NA NA
j06 1 1 1 1 1 NA NA NA NA NA NA NA NA
j07 1 0 NA 0 0 0 0 0 0 0 0 0 NA
j08 1 NA 0 0 0 0 0 0 0 0 0 0 NA
j09 1 NA 0 NA NA 1 NA NA NA NA 1 1 NA
j10 1 NA 1 1 1 1 1 1 1 1 1 0 NA
原始数据集结构如下，
structure(list(id = c("j01","j02","j03","j04","j05","j06","j07","j08","j09","j10"),pr_1 = c(1,1,1),pr_2 = c(1,NA,NA),pr_3 = c(1,pr_4 = c(1,pr_5 = c(1,pr_6 = c(0,pr_7 = c(0,pr_8 = c(0,pr_9 = c(0,pr_10 = c(0,pr_11 = c(0,pr_12 = c(0,0)),row.names = c(NA,-10L),class = c("tbl_df","tbl","data.frame"))->df
解决方法
您可以添加所有 pr 列的行列总和。
df$try <- rowSums(df[-1]) + 1
# id pr_1 pr_2 pr_3 pr_4 pr_5 pr_6 pr_7 pr_8 pr_9 pr_10 pr_11 pr_12 try
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 j01 1 1 1 1 1 0 0 0 0 0 0 0 6
# 2 j02 1 1 1 0 0 0 0 0 0 0 0 0 4
3 3 j03 1 0 0 0 0 0 0 0 0 0 0 0 2
# 4 j04 1 1 1 1 1 1 1 1 1 1 1 1 13
# 5 j05 1 1 1 1 1 1 1 1 NA 1 1 0 NA
# 6 j06 1 1 1 NA NA NA NA NA NA NA NA NA NA
# 7 j07 1 0 0 0 0 0 0 0 0 0 0 0 2
# 8 j08 1 NA 0 0 0 0 0 0 0 0 0 0 NA
# 9 j09 1 NA NA NA 1 NA NA NA 1 NA NA NA NA
#10 j10 1 NA 1 1 1 1 1 1 1 1 1 0 NA
或者使用 dplyr :
library(dplyr)
df %>% mutate(try = rowSums(select(.,starts_with('pr'))) + 1)
我相信这会给你“尝试”列rowSums(apply(df[,2:13],2,function(x) (x == 1))) + 1。一般的想法是按列检查元素是否等于1，然后按行求和。请注意，您提供的数据集中的 pr_3 列与您上面显示的列不同。为了获得您想要的相同结果，我认为这是一个错字并将其从 pr_3 = c(1,1,NA,1) 更改为 pr_3 = c(1,1)。
您也可以使用此代码。但是我相信当有 101111 这样的模式时，您想拥有 NA 吗？无论如何，下面的代码不能那样工作，并且仍然计算 1。
df %>%
tidyr::pivot_longer(-id) %>%
dplyr::group_by(id) %>%
dplyr::mutate(try = sum(value) + 1) %>%
dplyr::ungroup() %>%
tidyr::pivot_wider(names_from = name,values_from = value)