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在R中使用apply创建假人

apply r

1

Student · 技术社区 · 7 年前

我有关于个人不同学习策略的数据(存储在标有StrategyA、StrategyB、StrategyC的列中)。策略编码为1-15。我想为每个策略(例如策略1、策略2等)创建一个虚拟对象,因为每个学生最多可以列出3个策略。

   ID = c(1, 2, 3, 4, 5)
   Strategy_A = c(10, 12, 13, 1, 2)
   Strategy_B = c(1, 2, 1, 4, 5)
   Strategy_C = c(2, 3, 6, 8, 15)
   all = data.frame(ID, Strategy_A, Strategy_B, Strategy_C)

我考虑过使用apply并创建一个链接到fastDummies包的函数。

     dummies = function(x){
     dummy_cols(x)
     }

    new = apply(all [,-1], 2, dummies)
    new = as.data.frame(new)

2 回复 | 直到 7 年前

1

nghauran 7 年前

all dummy.data.frame() 从…起 dummies dummy_cols() 从…起 fastDummies aggregate 每 ID

all <- data.frame(ID = rep(all$ID, 3),
                  Strategy = c(all$Strategy_A, all$Strategy_B, all$Strategy_C)) # data frame "all" with one column Strategy
library(dummies)
all <- dummy.data.frame(all, "Strategy") # or fastDummies::dummy_cols(all, "Strategy")
aggregate(. ~ ID, all, sum) # since strategies are now dummies, the sum will always be 0 or 1
# output
  ID Strategy1 Strategy2 Strategy3 Strategy4 Strategy5 Strategy6 Strategy8 Strategy10 Strategy12 Strategy13 Strategy15
1  1         1         1         0         0         0         0         0          1          0          0          0
2  2         0         1         1         0         0         0         0          0          1          0          0
3  3         1         0         0         0         0         1         0          0          0          1          0
4  4         1         0         0         1         0         0         1          0          0          0          0
5  5         0         1         0         0         1         0         0          0          0          0          1

2

0

Darren Tsai 7 年前

我提供了一个 tidyverse 是的。

library(tidyverse)

new <- all %>% gather(select = -ID) %>%
               mutate(key = NULL, num = 1) %>%
               spread(value, num)

#   ID  1  2  3  4  5  6  8 10 12 13 15
# 1  1  1  1 NA NA NA NA NA  1 NA NA NA
# 2  2 NA  1  1 NA NA NA NA NA  1 NA NA
# 3  3  1 NA NA NA NA  1 NA NA NA  1 NA
# 4  4  1 NA NA  1 NA NA  1 NA NA NA NA
# 5  5 NA  1 NA NA  1 NA NA NA NA NA  1

new[is.na(new)] <- 0
new

#   ID 1 2 3 4 5 6 8 10 12 13 15
# 1  1 1 1 0 0 0 0 0  1  0  0  0
# 2  2 0 1 1 0 0 0 0  0  1  0  0
# 3  3 1 0 0 0 0 1 0  0  0  1  0
# 4  4 1 0 0 1 0 0 1  0  0  0  0
# 5  5 0 1 0 0 1 0 0  0  0  0  1