更改列表理解以在不带方括号和引号的打印语句中显示结果

Question

我正在尝试做一个简单的数学运算，这应该是非常简单易行的。

I24 = '4.64%'
J24 = 96.8
K24 = 96.8

我在我的 print 语句中使用了两个列表理解，不确定我是否应该对这样的东西使用列表理解

['control' if i < 50 else 'variant' for i in [J24]] ## if J24 is less than 50 then 'control' else 'variant'.. as you see I'm taking J24 into a list

# if K24 < 75 then 'No evidence of difference' if K24 >= 85 then 'Meaningful difference', if K24 between 75 & 84 then 'Marginal difference'

['No evidence of difference' if x < 75 else 'Meaningful difference' if x >= 85 else 'Marginal difference' for x in [K24]]

当我打印语句时--

print("V to Qc uplift "+I24+' ('+str(K24)+'%'+' probability of '+ str(['control' if i < 50 else 'variant' for i in [J24]]) + ' being better). '+str(['No evidence of difference' if x < 75 else 'Meaningful difference' if x >= 85 else 'Marginal difference' for x in [K24]]))

我的输出看起来像这样

V to Qc uplift 4.64% (96.8% probability of ['variant'] being better). ['Meaningful difference']

我希望我的输出看起来没有方括号和引号

像这样：

V to Qc uplift 4.64% (96.8% probability of variant being better). Meaningful difference

我已经使用了列表理解，但我不确定这是否是最佳解决方案。非常感谢你的帮助 ！ :)

Answer 1

试试这个：

print("V to Qc uplift "+I24+' ('+str(K24)+'%'+' probability of '+ str('control' if J24 < 50 else 'variant') + ' being better). '+str('No evidence of difference' if K24 < 75 else 'Meaningful difference' if K24 >= 85 else 'Marginal difference'))

不要执行 for 循环，而是执行 if 语句。