初学者想知道他的代码是否是'Pythonic'

Question

这是我在python中编写的第一件事.我来自Java背景.我不想只学习如何用Python语法编写java代码.我想学习如何用pythonic范例编程.

你能不能评论我如何使下面的代码更pythonic？

from math import sqrt

# recursively computes the factors of a number
def factors(num):
    factorList = []
    numroot = int(sqrt(num)) + 1
    numleft = num
    # brute force divide the number until you find a factor
    for i in range(2, numroot):
        if num % i == 0:
            # if we found a factor, add it to the list and compute the remainder
            factorList.append(i)
            numleft = num / i
            break
    # if we didn't find a factor, get out of here!
    if numleft == num: 
        factorList.append(num)
        return factorList
    # now recursively find the rest of the factors
    restFactors = factors(numleft)
    factorList.extend(restFactors)

    return factorList

# grabs  all of the twos in the list and puts them into 2 ^ x form
def transformFactorList(factorList):
    num2s = 0
    # remove all twos, counting them as we go
    while 2 in factorList:
        factorList.remove(2)
        num2s += 1
    # simply return the list with the 2's back in the right spot
    if num2s == 0: return factorList
    if num2s == 1:
        factorList.insert(0, 2)
        return factorList
    factorList.insert(0, '2 ^ ' + str(num2s))
    return factorList

print transformFactorList(factors(#some number))

Answer 1

大卫·古德(David Goodger)在这里有一本名为"像Python一样的代码"的优秀入门书.重新命名(引用)该文本的几件事:

• joined_lower 用于功能,方法,属性

• joined_lower 或常量的ALL_CAPS

• StudlyCaps 对于课程

• camelCase 只是为了符合已有的惯例

Answer 2

只需使用'import math'和'math.sqrt()'代替'from math import sqrt'和'sqrt()'; 你只是导入'sqrt'就不会赢得任何东西,并且代码很快会因为太多的导入而变得笨拙.此外,当你大量使用from-import时,reload()和mock for out等测试会更快地破解.

divmod()函数是执行除法和模数的便捷方式.您可以使用/ else而不是对numleft进行单独检查.您的因子功能是发电机的自然候选者.另一个答案中已经提到了xrange().这就是这样做的全部:

import math

# recursively computes the factors of a number as a generator
def factors(num):
    numroot = int(math.sqrt(num)) + 1
    # brute force divide the number until you find a factor
    for i in xrange(2, numroot):
        divider, remainder = divmod(num, i)
        if not remainder:
            # if we found a factor, add it to the list and compute the
            # remainder
            yield i
            break
    else:
    # if we didn't find a factor, get out of here!
        yield num
        return
    # now recursively find the rest of the factors
    for factor in factors(divider):
        yield factor

使用生成器意味着您只能迭代结果一次; 如果您只想要一个列表(就像在translateFactorsList中那样),则必须将调用包装在lists()中的factors()中.

Answer 3

您可能想要看的另一件事是文档字符串.例如,此函数的注释:

# recursively computes the factors of a number
def factors(num):

可以转换成这个:

def factors(num):
    """ recursively computes the factors of a number"""

这样做并不是真的100%必要,但如果你开始使用pydoc的东西,这是一个很好的习惯.

你也可以这样做:

docstring.py

"""This is a docstring"""

在命令行:

>>> import docstring
>>> help(docstring)

结果:

Help on module docstring:

NAME
    docstring - This is a docstring

FILE
    /Users/jason/docstring.py

Answer 4

一些评论:

1. 我会替换range()为xrange(); 当你调用时range(),它会同时分配整个范围,而当你迭代时xrange(),它会一次返回一个结果,从而节省内存.
2. 不要将条件后的表达式放在同一行(if num2s -- 0: return factorList)上.它使得一眼就能看到它正在做什么变得更加困难(这是一个障碍).
3. 不要害怕使用模块.该[sympy][1]模块已经具有计算因子的代码,这可以通过消除大部分代码来简化代码.
4. Python的字符串格式化简单而有效.

例如:

factorList.insert(0, '2 ^ ' + str(num2s))

可以改为

factorlist.insert(0, '2 ^ %s' % num2s)

总而言之,我发现你的代码并不是广泛的非pythonic.只要确保你想要使用地板划分,因为这是默认情况下会发生的整数值.否则,您需要修复除法运算符:

from __future__ import division

有时令人沮丧的语言警告.

Answer 5

这是我突然想到的：

def transformFactorList(factorList):
    oldsize = len(factorList)
    factorList = [f for f in factorList if f != 2]
    num2s = oldsize - len(factorList)
    if num2s == 0:
        return []
    if num2s == 1:
        return [2]+factorList
     return ['2 ^ %s' % num2s] + [factorList]

这种形式[f for f in factorList if f != 2]称为列表理解。