Question

我有一个带有两个布尔列A和B的data.table。我想添加一个新的布尔行C，它依赖于A和B，但是我遇到了麻烦＆＃39; look＆＃39;在之前和上升的行中。

我想按如下方式定义C.如果有一个A = 1的行，并且在三行的范围内至少有一个B = 1那么我希望C在所有的A = 1和C = 0的行上变成C = 1范围内的其他行。否则C应为C = B.

在两个范围重叠并且两者都包含B = 1的情况下，C在两个行上应该变为C = 1，其中A = 1且C = 0。有关更多说明：

df <- data.table(A=c(0,0,0,1,0,0,0,0,0,0,0,1,1,0,0), 
                 B=c(0,1,0,0,0,1,0,1,1,0,0,0,0,0,1))

    A B                                        A B C
1:  0 0 #                                  1:  0 0 0
2:  0 1 #                                  2:  0 1 0
3:  0 0 #                                  3:  0 0 0
4:  1 0 # range of three                   4:  1 0 1
5:  0 0 #                                  5:  0 0 0
6:  0 1 #                                  6:  0 1 0
7:  0 0 #                                  7:  0 0 0
8:  0 1                                    8:  0 1 1 # C = B
9:  0 1 #                                  9:  0 1 0
10: 0 0 ##                                 10: 0 0 0
11: 0 0 ##                                 11: 0 0 0
12: 1 0 ## overlapping range of three      12: 1 0 1
13: 1 0 ##                                 13: 1 0 1
14: 0 0 ##                                 14: 0 0 0
15: 0 1 ##                                 15: 0 1 0

我该如何做到这一点，我对这一点毫无头绪。

Answer 1

# Find ranges where A == 1
ind <- lapply(which(df$A == 1)
              , function(i){s <- i + -3:3; s[s %in% seq(nrow(df))]})
# Remove ranges with no B == 1
good <- sapply(ind, function(i) df[i, any(B == 1)])
ind  <- unique(unlist(ind[good]))
# Assign C as described
df[, C := B]
df[ind, C := as.numeric(A == 1)]
df
#     A B C
#  1: 0 0 0
#  2: 0 1 0
#  3: 0 0 0
#  4: 1 0 1
#  5: 0 0 0
#  6: 0 1 0
#  7: 0 0 0
#  8: 0 1 1
#  9: 0 1 0
# 10: 0 0 0
# 11: 0 0 0
# 12: 1 0 1
# 13: 1 0 1
# 14: 0 0 0
# 15: 0 1 0

以下使用的数据。我更改了您的df定义以匹配显示的df

df <- data.table(A=c(0,0,0,1,0,0,0,0,0,0,0,0,1,0,0), 
                 B=c(0,1,0,0,0,1,0,1,1,0,0,0,0,0,0))

df[12, A := 1]
df[15, B := 1]

df

#     A B
#  1: 0 0
#  2: 0 1
#  3: 0 0
#  4: 1 0
#  5: 0 0
#  6: 0 1
#  7: 0 0
#  8: 0 1
#  9: 0 1
# 10: 0 0
# 11: 0 0
# 12: 1 0
# 13: 1 0
# 14: 0 0
# 15: 0 1

Answer 2

这是一个基于tidyverse套件的解决方案：

我定义了2个临时变量 - A1确定{row 1：row + 3}窗口中的A = 1是否在任何位置。 C1会在窗口的任何位置测试A = 1和B = 1。

library(tidyverse)
df %>% 
  mutate(
    A1 = (cumsum(lead(A, 3, default = 0)) - cumsum(dplyr::lag(A, 4, default = 0)) > 0),
    C1 = (A & dplyr::lead(cumsum(B), n = 3, default = 0) - dplyr::lag(cumsum(B), n = 4, default = 0)) * 1,
    C = ifelse(!A1, B, C1)
  ) %>%
  select(-A1, -C1)

R根据不同列中某个范围内的值添加新列

2 个答案: