2016-09-24 69 views
2

我有許多條件列表,我想評估它們的組合,然後我想獲取這些邏輯值的二進制值(True = 1,假= 0)。條件本身可能隨着我的項目進展而改變或增加,所以我希望在腳本中有一個可以改變這些條件語句的位置,而腳本的其餘部分保持不變。從語句向量中獲取邏輯值的數據框

這裏是一個簡化的,可重複的例子:

# get the data 
df <- data.frame(id = c(1,2,3,4,5), x = c(11,4,8,9,12), y = c(0.5,0.9,0.11,0.6, 0.5)) 

# name and define the conditions 
names1 <- c("above2","above5") 
conditions1 <- c("df$x > 2", "df$x >5") 

names2 <- c("belowpt6", "belowpt4") 
conditions2 <- c("df$y < 0.6", "df$y < 0.4") 

# create an object that contains the unique combinations of these conditions and their names, to be used for labeling columns later 

names_combinations <- as.vector(t(outer(names1, names2, paste, sep="_"))) 

condition_combinations <- as.vector(t(outer(conditions1, conditions2, paste, sep=" & "))) 

# create a dataframe of the logical values of these conditions 

condition_combinations_logical <- ????? # This is where I need help 

# lapply to get binary values from these logical vectors 

df[paste0("var_",names_combinations] <- +(condition_combinations_logical) 

獲得輸出,可能看起來像:

-id -- | -x -- | -y -- | -var_above2_belowpt6 -- | -var_above2_belowpt4 -- | etc. 
1  | 11 | 0.5 | 1      | 0      | 
2  | 4 | 0.9 | 0      | 0      | 
3  | 8 | 0.11 | 1      | 1      | 
etc. .... 

回答

1

貌似可怕的eval(parse())它(很難想象的要容易得多辦法 ...)。然後使用storage.mode()<-從邏輯轉換爲整數...

res <- sapply(condition_combinations,function(x) eval(parse(text=x))) 
storage.mode(res) <- "integer" 
+0

我認爲在這裏使用eval(parse())沒有任何問題。這是它打算做的。 – dracodoc