超簡易R for循環不工作

Question

我想不出這裏發生了什麼（R語言）：

a = c(1, 1, 1, 1, 0, 0, 0, 0, 0) 
space = combn(a,2) 
b = 0 
for(j in ncol(space)){ 
if(space[1, j] == space[2, j]){ 
    b = b + 1 
    } 
} 

我得到B = 1，這不應該是1¿什麼想法？

提前致謝！

Answer 1

我們可以做到這一點沒有任何環

combn(a, 2, FUN = function(x) +(x[1]==x[2])) 
#[1] 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 

---

如果上面的輸出是預計，隨着for一層的方法是將初始化「B」與length等於「樣品的ncol '然後通過列的順序循環‘空間’

b <- numeric(ncol(space)) 
for(j in 1:ncol(space)){ 
    if(space[1, j] == space[2, j]){ 
    b[j] = b[j] + 1 
    } 
}  

b 
#[1] 1 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 

注：在OP的代碼for LO op是j in ncol(space)，它的長度是1，而不是循環遍歷序列。

---

如果我們想去的地方都值是1或兩者都是0

sum(colSums(space)==2) 
#[1] 6 
sum(!colSums(space)) 
#[1] 10 

或者使用table

tbl <- table(combn(a, 2, FUN = sum)) 
tbl[names(tbl)!=1] 

# 0 2 
#10 6 

使用OP的代碼列數（改爲1:ncol(space)）

b <- 0 
for(j in 1:ncol(space)){ 
    #Note that here we are not differentiating whether both 
    #are 0 or both are 1 
    if(space[1, j] == space[2, j]){ 
    b = b + 1 
     } 
    } 
b 
#[1] 16 

但是，假設我們做

b <- 0 
for(j in 1:ncol(space)){ 
    if(space[1, j] == 1 & space[2,j]==1){ 
    b = b + 1 
    } 
    } 
b 
#[1] 6 

Answer 2

可以在各種條件

# Both the rows having value 1 
> which(space[1,] == 1 & space[2,] == 1) 
[1] 1 2 3 9 10 16 
# Both the rows having value 0 
> which(space[1,] == 0 & space[2,] == 0) 
[1] 27 28 29 30 31 32 33 34 35 36 
# Row 1 having value 1 and row 2 having value 0 
> which(space[1,] == 1 & space[2,] == 0) 
[1] 4 5 6 7 8 11 12 13 14 15 17 18 19 20 21 22 23 24 25 26 
# ROw 1 having value 0 and row 2 having value 1 
> which(space[1,] == 0 & space[2,] == 1) 
integer(0) 

# Total number obtained from each case above: 
> length(which(space[1,] == 1 & space[2,] == 1)) 
[1] 6 

> length(which(space[1,] == 0 & space[2,] == 0)) 
[1] 10 

> length(which(space[1,] == 1 & space[2,] == 0)) 
[1] 20 

> length(which(space[1,] == 0 & space[2,] == 1)) 
[1] 0