Question

I have a data vector with 1024 values and need to count the number of negative entries. Is there an elegant way to do this without looping and checking if an element is <0 and incrementing a counter?

Answer 1

页: 1 你在这里的回答只是说

 sum(  x < 0  )

由于病媒化,这些产品发挥了作用。 <代码>x < 0 表述了青).体的载体,sum(<>>/code>> 可操作(通过将ole改成标准0/1数值)。

Answer 2

对这一问题有一个很好的答案:Steve Lianoglou 。如何在我的数据框架内确定在任何一栏中具有负值的浏览量?

让我稍加补充一下他的法典(4点)。

你们有数据。类似:

df <- data.frame(a = 1:10, b = c(1:3,-4, 5:10), c = c(-1, 2:10))

这将使你们重新成为一流具有消极价值的风气病媒:
```
has.neg <- apply(df, 1, function(row) any(row < 0))
```
此处为负数指数:
```
which(has.neg)
```
这里有负数的要素:
```
length(which(has.neg))
```

Answer 3

The above solutions prescribed need to be tweaked in-order to apply this for a df. The below command helps get the count of negative or any other symbolic logical relationship. Suppose you have a dataframe:

df <- data.frame(x=c(2,5,-10,NA,7), y=c(81,-1001,-1,NA,-991))

在<条码>x中获取负面记录的顺序:

nrow(df[df$x<0,])

友情链接