如何交叉制表R中两个变量的计数=1？

我在R中有一个数据帧：

id var1 var2 var3 var4 var5
1   0    1    0    1    0
2   1    1    1    1    0
3   0    1    1    0    1
4   0    0    0    1    0
5   0    0    0    0    0

var1-5具有0、1或。价值观我对计算varX=1的共现频率感兴趣：

     var1 var2 var3 var4 var5
var1  0    1    1    1    0
var2  1    0    2    2    1
var3  1    2    0    1    1
var4  1    2    1    1    0
var5  0    1    1    0    0

var1-var1为0，因为没有受试者在所有其他变量中的var1=1和0。这也可能是,，我对其他频率更感兴趣。 var2-var1是1，因为只有一个受试者的var2和var1都等于1（id=2）

有什么想法吗

///

编辑：这是来自

dput（头（s1））的输出。

：

---

我不明白你想要的结果的逻辑。共现事件频率的弦图。你能举一个类似的小例子来说明你想要的输出是什么样的吗？我想你想要

`diag@27ν9它导致
错误：意外“，”in“诊断

    structure(list(NBOMe_lifetime = c(0, 0, 0, 0, 0, 0), CB_lifetime = c(0, 
    0, 0, 0, 0, 0), HOMET_lifetime = c(0, 0, 0, 0, 0, 0), DMT_lifetime = c(0, 
    0, 0, 0, 0, 0), aPVP_lifetime = c(0, 0, 0, 0, 0, 0), Amph_lifetime = c(0, 
    0, 0, 0, 0, 0), BZP_lifetime = c(0, 0, 0, 0, 0, 0), THC_lifetime = c(1, 
    1, 1, 1, 1, 1), Coca_lifetime = c(0, 0, 0, 0, 0, 0), Ero_lifetime = c(0, 
    0, 0, 0, 0, 0), GHB_lifetime = c(0, 0, 0, 0, 0, 0), spice_lifetime = c(0, 
    0, 0, 0, 0, 0), JWH250_lifetime = c(0, 0, 0, 0, 0, 0), Keta_lifetime = c(0, 
    0, 0, 0, 0, 0), Kath_lifetime = c(0, 0, 0, 0, 0, 0), Kratom_lifetime = c(0, 
    0, 0, 0, 0, 0), LSD_lifetime = c(0, 0, 0, 0, 0, 0), MDMA_lifetime = c(1, 
    0, 1, 0, 0, 0), MDPV_lifetime = c(0, 0, 0, 0, 0, 0), Mephe_lifetime = c(0, 
    0, 0, 0, 0, 0), Meth_lifetime = c(0, 0, 0, 0, 0, 0), MTX_lifetime = c(0, 
    0, 0, 0, 0, 0), Ocfenta_lifetime = c(0, 0, 0, 0, 0, 0), Oxicod_lifetime = c(0, 
    0, 0, 0, 0, 0), PCP_lifetime = c(0, 0, 0, 0, 0, 0), Popper_lifetime = c(1, 
    0, 1, 0, 1, 0), Rital_lifetime = c(0, 0, 0, 0, 0, 0), Salvia_lifetime = c(0, 
    0, 0, 0, 0, 0)), row.names = c(NA, 6L), class = "data.frame")