3

我怎样才能摆脱这段代码中的过度重复？

代码： http: //pastebin.com/13e2nWM9

该程序根据用户提供的数据计算运动方程（SUVAT 方程）。

我指的部分是：

while True:
        a = raw_input("What is the value of a? Make sure it is in standard units, however, do not include the unit.")
        try:
            float(a)
            print "a is " + str(a) + " ms^-2"
            break
        except:
            print"You must enter a number. Don't include units!"

这会重复很多次，除了变量“a”和重复块时改变的单位。

非常感谢。

4

4 回答 4

3

这是一种选择，将以下函数定义放在模块的顶部：

def get_float(name, units):
    prompt = "What is the value of {0}? Make sure it is in standard units, however, do not include the unit.".format(name)
    while True:
        val = raw_input(prompt)
        try:
            val = float(val)
            print '{0} is {1} {2}'.format(name, val, units)
            return val
        except Exception:
            print "You must enter a number. Don't include units!"

这是一个如何使用它的示例，以下代码可以替换从第 72 行到第 100 行的所有内容：

name_units_pairs = [('v', 'ms^-1'), ('a', 'ms^-2'), ('t', 's'),]
vals = {}
for name, units in name_units_pairs:
    vals[name] = get_float(name, units)
u = vals['v'] - vals['a'] * vals['t']
于 2013-04-11T23:45:10.023 回答
1

不要使用原始异常，并封装测试，仅此而已。所以而不是

while True:
        a = raw_input("What is the value of a? Make sure it is in standard units, however, do not include the unit.")
        try:
            float(a)
            print "a is " + str(a) + " ms^-2"
            break
        except:
            print"You must enter a number. Don't include units!"

做

a = my_input(valuename='a', unit='m^-2', unitformat=float)

并在 my_input 中进行测试（和提示）。

my_input 可能类似于：

def my_input(valuename, unit, unitformat=float):
  while True:
    val = raw_input("What is the value of %s? Make sure it is in standard units"+
                    "(%s), however, do not include the unit." % (valuename, unit))
    try:
        unitformat(val)
    except ValueError:
        print"You must enter a number. Don't include units!"
    else:
        print "%s is %s %s" % (valuename, val, unit)
        return val
于 2013-04-11T23:43:37.060 回答
0

我看不出这段代码有任何需要，或者事实上根本没有任何用途，以提高效率。这里的限制因素是控制台输出的速度。

但...

如果它为您节省的额外几纳秒真的非常重要，那么您可以加快速度的最大方法（我认为，尽管它实际上可能会因为解释而减慢速度）是以不同的方式单独输出打印的内容print 语句，因此它不必在内存中构造 2 个额外的字符串来表示"a is " + str(a)and的值"a is " + str(a) + " ms^-2"。为了再节省几纳秒，请用一种可以编译为机器代码的语言编写程序。

或者，如果您的意思是使用更少的代码来做同样的事情，请封装输入，就像其他人所说的那样。

于 2013-04-11T23:41:54.060 回答
0

您可以使用循环来自动化输入，将值存储在字典中，然后在最后进行所有数学运算。

import math

units = {'s': 'm', 'u': 'ms^-1', 'v': 'ms^-1', 'a': 'ms^-2', 't': 's'}
variables = ['s', 'u', 'v', 'a', 't']
suvat = dict((k, None) for k in variables)
input_variables = []

def pick_from_list(prompt, L):
    while True:
        pick = raw_input(prompt % str(L))
        if pick in L:
            print "You have chosen", pick
            return pick
        print """Sorry, I didn't understand that, try again. Make sure your spelling is
            correct (Case Sensitive), and that you did not inlcude the quotation marks."""

aim = pick_from_list("""Welcome to Mattin's SUVAT Simulator! Choose the value you are
                    trying to find. You can pick from variables %s""",
                     variables)

for order in 'first', 'second', 'third':
    prompt = 'Please choose which variable to input %s. You can pick from %%s' % order
    choices = [k for k in variables if not (k is aim or k in input_variables)]
    input_variables.append(pick_from_list(prompt, choices))

for v in input_variables:
    while True:
        raw_input("""What is the value of %s? Make sure it is in standard
units, however, do not include the unit.""" % v)
        try:
            suvat[v] = float(val)
            print "%s is %s%s" % (v, val, units[v])
            break
        except:
            print "You must enter a number. Don't include units!"

# for readability of the maths, turn dictionary into variables
s, u, v, a, t = [suvat[k] for k in variables]

# (not tested)
if s is None:
    if t is None:
        t = (v - u) / a
    if v is None:
        s = u * t + .5 * a * t * t
    elif a is None:
        s = .5 * (u + v) * t
    else:
        s = v * t - .5 * a * t * t
if v is None:
    if u is None:
        u = s / t - .5 * a * t * t
    if t is None:
        v = math.sqrt(u * u + 2 * a * s)
    else:
        v = 2 * s / t - u
if u is None:
    if a is None:
        a = 2 * v / t - s / (t * t)
    u = math.sqrt(v * v - 2 * a * s)
if a is None:
    a = (v * v - u * u) / (2 * s)
if t is None:
    t = (v - u) / a

# turn the set of variables back into a dictionary
solutions = dict(zip(variables, [s, u, v, a, t]))
print 'Solution: %s=%s%s' % (aim, solutions[aim], units[aim])
于 2013-04-12T01:54:14.583 回答