Question

在最低和最高限值可能无效的情况下,Im 寻求一种简单的盘点浮标方式。

该法典是:

    tval = float(-b - discriminant) / float (2*a)
    if tval >= tmin and tval <= tmax:
        return tval 

    tval = float(-b + discriminant) / float (2*a)
    if tval >= tmin and tval <= tmax:
        return tval

    # Neither solution was within the acceptable range.
    return None

然而,这完全不能处理无母体或最高权的情况(这应理解为没有最低或最高标准)。

迄今为止,我最能前来的就是:

    tval = float(-b - discriminant) / float (2*a)
    if (tmin == None or tval >= tmin) and (tmax == None or tval <= tmax):
        return tval 

    tval = float(-b + discriminant) / float (2*a)
    if (tmin == None or tval >= tmin) and (tmax == None or tval <= tmax):
        return tval

    # Neither solution was within the acceptable range.
    return None

我始终认为,必须找到更好的(更清晰、更可读)方式来写。任何想法?

Answer 1

利用点击的答复中的INF

if tmin is None: tmin = -INF
if tmax is None: tmax = +INF

tval = float(-b - discriminant) / float (2*a)
if tmin <= tval <= tmax:
    return tval 

tval = float(-b + discriminant) / float (2*a)
if tmin <= tval <= tmax:
    return tval

# Neither solution was within the acceptable range.
return None

Answer 2

可读性:

def inrange(x, min, max):
    return (min is None or min <= x) and (max is None or max >= x)

tval = float(-b - discriminant) / float (2*a)
if inrange(tval, tmin, tmax):
    return tval 

tval = float(-b + discriminant) / float (2*a)
if inrange(tval, tmin, tmax):
    return tval 

# Neither solution was within the acceptable range.
return None

必须有一套界定“<条码>的模块。但是,我没有发现它(也看不到它,但:)

Answer 3

第一,一些结构:我们需要固定的浮动。

INF = float(1e3000)

或

INF = float( inf )  # Python 2.6+

First option can be considered p或table f或 practical purposes; just use some really huge value which is guaranteed to be outside the range representable by your platf或m s floating point type. Second option is "truly" p或table, but requires Python 2.6 或 newer.

现在,你的条件可以这样写(edit:只有tmin和tmax。不能是零!

if (tmin 或 -INF) <= tval <= (tmax 或 +INF) : return tval

Edit

I made a crude err或 by missing that 0.0 is a legitimate value f或 tmin 或 tmax. Thanks to Roger Pate f或 noticing.

Answer 4

采用zz式回答(在使用<0> <<>>>>>>>/代码>时,以一定失败的方式):

def coalesce(*values):
  for v in values:
    if v is not None:
      return v

if coalesce(tmin, -INF) <= tval <= coalesce(tmax, INF):
  return tval

然而,我很清楚:

if ((tmin is None or tmin <= tval) and
    (tmax is None or tval <= tmax)):
   return tval

Edit

友情链接