首页
学习
活动
专区
工具
TVP
发布
社区首页 >问答首页 >Pythonic式的修改列表中包含特定数字(而不仅仅是字符)的元素的方法

Pythonic式的修改列表中包含特定数字(而不仅仅是字符)的元素的方法
EN

Stack Overflow用户
提问于 2018-12-05 03:58:04
回答 1查看 38关注 0票数 2

我正在写一个' simplify‘函数来简化多项式,这样simplify("2xy-yx")就可以返回"xy"simplify("-a+5ab+3a-c-2a")就可以返回"-c+5ab",依此类推。

在这个阶段,我合并了相同的单项式,但某些单项式的系数将为+1-1。我想只将它们更改为+-。但我不能直接删除1,否则值为+12-18的系数将分别更改为+2-8

例如

input = '3xy+y-2x+2xy'

到目前为止,我的流程给了我:

Var = ['xy', 'y', 'x']
Coe = ['+5', '+1', '-2']

然后我得到了一个中间结果:IR = ['+5xy', '+1y', '-2x']

我目前的解决方案是:

[(e[0]+e[2:]) if (e[1]== '1' and e[2].isalpha() ) else e for e in IR ]

它现在似乎为我做了这项工作,但看起来很笨拙。我想知道是否有更干净、更简洁的方法来实现同样的目标。最好不要使用正则表达式。

谢谢。

编辑:我为这个问题编写的代码。我还在调试它。

def simplify(poly):

    #If the first coefficient is positive, then I add a + for calculation later on
    if poly[0] !='-':
        poly = '+'+poly
    L = list(poly)

    #Put each monomial in a list as an element
    Temp, e = [], ''
    for i in L:
        if i != '+' and i != '-':
            e += i
        else:
            Temp.append(e)
            e = ''
            e += i

    #The last one will be left out and the first one will be None
    Temp.append(e)
    Temp.pop(0)

    #If the monomial only has a + or - in front of it, then give it a '1' so it's easier for calculation
    SortAndGiveOne = [''.join(e[0] + '1' + e[1:]) if not e[1].isdigit() else e for e in Temp]

    #Try to get letters from each element of the list
    Var = [''.join(sorted(i for i in e if i.isalpha())) for e in SortAndGiveOne]
    #Try to get coefficients from each element of the list
    Coe = [''.join(i for i in e if not i.isalpha()) for e in SortAndGiveOne]

    #Calculation of equivalent monomials
    newvar = []
    newcoe = []
    for va, co in zip(Var, Coe):
        try:
            ind = newvar.index(va)
            newcoe[ind] = str(int(newcoe[ind]) + int(co))
        except ValueError:
            newvar.append(va)
            newcoe.append(co)

    # Put the new lists together
    Put_t = list(map(lambda x,y : y + x, newvar, newcoe))

    # Add the plus sign as it will be gone if there was calculation involved.
    FinalSign = ['+'+ e if e[0] != '+' and e[0] != '-' else e for e in Put_t]

    #Delete the elements with 0 as coefficient
    Delete0 = [e for e in FinalSign if not e[1]=='0']

    #Change the +1 and -1 coefficients to + and - 
    Change1 = [(e[0]+e[2:]) if (e[1]== '1' and e[2].isalpha() ) else e for e in Delete0 ]

    #Sort the list based on length and then lexi order
    FS_sort = sorted(Change1, key = lambda s: (len(''.join(filter(str.isalpha, s))), (''.join(filter(str.isalpha, s)))))

    #Join together as a list
    JT = ''.join(FS_sort)

    #Delete leading plus sign
    if JT[0] == '+':
        JT = JT[1:]

    return JT
EN

回答 1

Stack Overflow用户

发布于 2018-12-05 04:17:21

看起来您只想将['5xy', '+1y', '-2x']修改为['5xy', '+y', '-2x']等。在本例中,我将使用正则表达式:

import re
newlist = [re.sub('(?<=[+-])1(?=[A-Za-z])', '', e) for e in your_list]

测试:

k = ['5xy', '+1y', '-2x', '-1xy', '+11xy']
newlist = [re.sub('(?<=^[+-])1(?=[A-Za-z])', '', e) for e in k]
>>> newlist
['5xy', '+y', '-2x', '-xy', '+11xy']

不过,这看起来并不比您的解决方案好多少。

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/53620531

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档