将数值转换为二进制(0/1) [英] Convert numeric values into binary (0/1)

问题描述

我有一个数据框，其中包含不同人的不同种类水果的计数.像下面一样

I have a data frame with counts of different kinds of fruits of different people. Like below

    apple  banana  orange
Tim     3       0       2
Tom     0       1       1
Bob     1       2       2

如何将其更改为二进制矩阵，即，如果一个人至少有一个水果，无论他有多少水果，那么我记录1，如果没有，则记录0.如下所示

How can I change it into a binary matrix, i.e. if a person has at least one fruit, no matter how many he has, then the I record 1, if not, record 0. Like below

    apple  banana  orange
Tim     1       0       1
Tom     0       1       1
Bob     1       1       1

推荐答案

这是您的data.frame:

x <- structure(list(apple = c(3L, 0L, 1L), banana = 0:2, orange = c(2L, 
1L, 2L)), .Names = c("apple", "banana", "orange"), class = "data.frame", row.names = c("Tim", 
"Tom", "Bob"))

还有您的矩阵:

as.matrix((x > 0) + 0)
    apple banana orange
Tim     1      0      1
Tom     0      1      1
Bob     1      1      1

更新

我不知道快速的就寝前发布会产生讨论，但是讨论本身很有趣，因此我想在这里总结一下:

Update

I had no idea that a quick pre-bedtime posting would generate any discussion, but the discussions themselves are quite interesting, so I wanted to summarize here:

我的直觉是简单地认为R中的TRUE和FALSE下方是数字1和0.如果尝试(不是很好的方法)检查​​等效性，例如1 == TRUE或0 == FALSE，则会得到TRUE.我的捷径方式(事实证明，比起 correct 花费的时间要长更多，或者至少是在概念上更正确的方式)是仅添加到我的TRUE和FALSE，因为我知道R会将逻辑向量强制转换为数字.

My instinct was to simply take the fact that underneath a TRUE and FALSE in R, are the numbers 1 and 0. If you try (a not so good way) to check for equivalence, such as 1 == TRUE or 0 == FALSE, you'll get TRUE. My shortcut way (which turns out to take more time than the correct, or at least more conceptually correct way) was to just add 0 to my TRUEs and FALSEs, since I know that R would coerce the logical vectors to numeric.

正确或至少更合适的方法是使用as.numeric转换输出(我认为这就是@ JoshO'Brien打算编写的内容).但是....不幸的是，这删除了输入的维属性，因此您需要将结果向量重新转换为矩阵，事实证明，该矩阵为 静止 的速度要比我在答案中添加0的速度快.

The correct, or at least, more appropriate way, would be to convert the output using as.numeric (I think that's what @JoshO'Brien intended to write). BUT.... unfortunately, that removes the dimensional attributes of the input, so you need to re-convert the resulting vector to a matrix, which, as it turns out, is still faster than adding 0 as I did in my answer.

阅读了评论和批评之后，我想我会再添加一个选项-使用apply遍历各列并使用as.numeric方法.与手动重新创建矩阵相比，它的速度要慢，但是比在逻辑比较中添加0的速度更快..

Having read the comments and criticisms, I thought I would add one more option---using apply to loop through the columns and use the as.numeric approach. That is slower than manually re-creating the matrix, but slightly faster than adding 0 to the logical comparison.

x <- data.frame(replicate(1e4,sample(0:1e3)))
library(rbenchmark)
benchmark(X1 = {
            x1 <- as.matrix((x > 0) + 0)
          },
          X2 = {
            x2 <- apply(x, 2, function(y) as.numeric(y > 0))
          },
          X3 = {
            x3 <- as.numeric(as.matrix(x) > 0)
            x3 <- matrix(x3, nrow = 1001)
          },
          X4 = {
            x4 <- ifelse(x > 0, 1, 0)
          },
          columns = c("test", "replications", "elapsed", 
                      "relative", "user.self"))
#   test replications elapsed relative user.self
# 1   X1          100 116.618    1.985   110.711
# 2   X2          100 105.026    1.788    94.070
# 3   X3          100  58.750    1.000    46.007
# 4   X4          100 382.410    6.509   311.567

all.equal(x1, x2, check.attributes=FALSE)
# [1] TRUE
all.equal(x1, x3, check.attributes=FALSE)
# [1] TRUE
all.equal(x1, x4, check.attributes=FALSE)
# [1] TRUE

谢谢大家的讨论！

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