Python浮点操作

如何在python中打印一个简单操作的全浮点结果？我的代码是：

if( int(array_Y[counter2]) == int(round(float(elem[0])))):
                if(int(round(float(elem[0]))) == 0):
                    negatiu_verdader += 1
                if(int(round(float(elem[0]))) == 1):
                    positiu_verdader += 1
                counter = counter + 1
            counter2 = counter2 + 1

        error =  float(1.0000- (1.0000 * counter / counter2))

        print "        ERROR!!!!!!!!!!!!!!!!!!!!!!!! :" + ("{0:.15f}".format(round(error,2)))

---

但是错误总是：0.420000000000000或0.230000000000000000000，但我希望错误是：0.4323321321232。

通过调用rounderror，2:

如果要显示更高的精度，请不要这样做：

>>> format(round(0.43233213213232, 2), '.15f')
'0.430000000000000'
>>> format(0.43233213213232, '.15f')
'0.432332132132320'

您在代码中做了大量冗余工作，请将其简化一点：

elem_rounded = int(round(float(elem[0])))
if int(array_Y[counter2]) == elem_rounded:
    if not elem_rounded:
        negatiu_verdader += 1
    elif elem_rounded == 1:
        positiu_verdader += 1
        counter += 1
        counter2 += 1

    error =  1.0 - (1.0 * counter / counter2)
    print "        ERROR!!!!!!!!!!!!!!!!!!!!!!!! :{0:.15f}".format(error)

---

Python if语句不需要括号；它们使你的代码在这里更难阅读。如果你要绕行并使用元素[0]，只需执行一次，而不是对所有If语句执行三次。

elem_rounded = int(round(float(elem[0])))
if int(array_Y[counter2]) == elem_rounded:
    if not elem_rounded:
        negatiu_verdader += 1
    elif elem_rounded == 1:
        positiu_verdader += 1
        counter += 1
        counter2 += 1

    error =  1.0 - (1.0 * counter / counter2)
    print "        ERROR!!!!!!!!!!!!!!!!!!!!!!!! :{0:.15f}".format(error)