for stylus for CSS选择器名称的循环

我正在使用手写笔和Im寻找功能以获得以下结果:

.post_1,.post_4,.post_7,.post_10,.post_13,.post_16,.post_19,.post_22 {
    left:0%;
}
.post_2,.post_5,.post_8,.post_11,.post_14,.post_17,.post_20,.post_23 {
    left:33.33%;
}
.post_3,.post_6,.post_9,.post_12,.post_15,.post_18,.post_21,.post_24 {
    left:66.66%;
}

这是我的尝试

post_class(a,b)
    for i in (0..a - 1)
        for u in (1..b)
            .post_{a * u + i - (a - 1)}
                left floor(i*(100/a)%,2)

post_class(3,8)// 3 columns, 8 rows

这给了我想要的CSS,但所有选择器都具有相同的属性(在这种情况下保留其值).

.post_1 {left: 0%;}
.post_4 {left: 0%;}
.post_7 {left: 0%;}
.post_10 {left: 0%;}
.post_13 {left: 0%;}
.post_16 {left: 0%;}
.post_19 {left: 0%;}
.post_22 {left: 0%;}
.post_2 {left: 33.33%;}
.post_5 {left: 33.33%;}
.post_8 {left: 33.33%;}
.post_11 {left: 33.33%;}
.post_14 {left: 33.33%;}
.post_17 {left: 33.33%;}
.post_20 {left: 33.33%;}
.post_23 {left: 33.33%;}
.post_3 {left: 66.66%;}
.post_6 {left: 66.66%;}
.post_9 {left: 66.66%;}
.post_12 {left: 66.66%;}
.post_15 {left: 66.66%;}
.post_18 {left: 66.66%;}
.post_21 {left: 66.66%;}
.post_24 {left: 66.66%;}

我只能想到循环选择器的名称,我试过没有运气(抛出错误).是否有可能实现我正在寻找的东西？(澄清一下,我正在寻找一种简化结果CSS的方法)