Python - list.remove(x) x 不在列表中

Question

我正在尝试在 Python 3.3 中创建一个简单的程序，该程序采用四个名称的列表，并将它们随机分配给列表中的另一个人。例如，如果名字是 John、Aaron、Lydia 和 Robin：

约翰首先选择了一个名字。他不能自己画；如果他这样做了，他就把它放回去，然后再画一次。假设约翰画了罗宾的名字。罗宾的名字将从池中删除。接下来轮到Aaron画了。他画约翰。约翰的名字被删除了。等等，直到所有名称都被分配。

我创建了一个包含四个名称的列表，并为每个名称分配了一个值 1-4。但是，我在从列表中删除时遇到了一个问题，说该值不存在。

list.remove(x) : x 不在列表中。

看起来像这样：

def drawNames():
    import random
    John=1
    Aaron=2
    Lydia=3
    Robin=4
    validNames=[John, Aaron, Lydia, Robin]
    nameDrawn=random.choice(validNames)
    def draw():
        nameDrawn=random.choice(validNames)
    #John's Draw:
    draw()
    if nameDrawn != 1:
        if nameDrawn == 2:
            print("John drew: Aaron")
            validNames.remove(2)
        elif nameDrawn == 3:
            print("John drew: Lydia")
            validNames.remove(3)
        elif nameDrawn == 4:
            print("John drew: Robin")
            validNames.remove(4)
    #Aaron's Draw:
    draw()
    if nameDrawn !=2:
        if nameDrawn ==1:
            print("Aaron drew: John")
            validNames.remove(1)
        elif nameDrawn ==3:
            print("Aaron drew: Lydia")
            validNames.remove(3)
        elif nameDrawn ==4:
            print("Aaron drew: Robin")
            validNames.remove(4)
    #Lydia's Draw:
    draw()
    if nameDrawn !=3:
        if nameDrawn ==1:
            print("Lydia drew: John")
            validNames.remove(1)
        elif nameDrawn ==2:
            print("Lydia drew: Aaron")
            validNames.remove(2)
        elif nameDrawn ==4:
            print("Lydia drew: Robin")
            validNames.remove(4)
    #Robin's Draw:
    draw()
    if nameDrawn !=4:
        if nameDrawn ==1:
            print("Robin drew: John")
            validNames.remove(1)
        elif nameDrawn ==2:
            print("Robin drew: Aaron")
            validNames.remove(2)
        elif nameDrawn ==3:
            print("Robin drew: Lydia")
            validNames.remove(3)
drawNames()

我也尝试过使用名称而不是数值，这产生了同样的错误。

我也觉得这是一个低效的方案。如果您有更好的建议，我将不胜感激。

Answer 1

使用下面这样的代码，您可能会获得更好的里程；它比您上面提供的名称更具可扩展性。

import copy
import random
validNames=["John", "Aaron", "Lydia", "Robin"]

def drawNames(namelist,currentname):
    '''
    namelist: list of names to draw from
    currentname: name of person doing the current draw
    '''
    draw_namelist = copy.copy(namelist) # make a copy to remove person drawing if needed
    if currentname in draw_namelist: # check if the person drawing is in the list
        draw_namelist.remove(currentname) # remove current name if in list
    try:
        drawn_name = random.choice(draw_namelist)
        namelist.remove(drawn_name)
        newnamelist = namelist
        print "Drew {}".format(drawn_name)
        print "New list: {}".format(newnamelist)
    except:
        print "Nobody for me to draw!"
        drawn_name=None
        newnamelist = namelist
    return drawn_name, newnamelist

然后可以按如下方式工作：

In [39]: newlist=["John", "Aaron", "Lydia", "Robin"]

In [40]: name,newlist = drawNames(newlist,"Lydia")
Drew Robin
New list: ['John', 'Aaron', 'Lydia']

In [41]: name,newlist = drawNames(newlist,"John")
Drew Aaron
New list: ['John', 'Lydia']

In [42]: name,newlist = drawNames(newlist,"Aaron")
Drew John
New list: ['Lydia']

In [43]: name,newlist = drawNames(newlist,"Robin")
Drew Lydia
New list: []

Answer 2

您必须调用实际元素，而不是 remove 中的索引。将数字替换为列表中存在的人的实际姓名，代码应按预期工作。

Answer 3

请注意，如果列表 len 为奇数，则可能存在无限循环

import random
from copy import copy
def f(name):
    x = random.choice(validNames)
    while x == name:
        x = random.choice(validNames)
    else:
        validNames.remove(x)
    return (name, x)

print map(f, copy(validNames))

>>> [('John', 'Robin'), ('Aaron', 'John'), ('Lydia', 'Aaron'), ('Robin', 'Lydia')]

Answer 4

你肯定在努力。

与其反复寻找列表并删除值，不如将列表打乱并读出对，如下所示：

from random import shuffle

def random_pairs(lst):
    shuffle(lst)
    return zip(lst[::2], lst[1::2])

def main():
    names = 'John,Aaron,Lydia,Robin'.split(',')
    print random_pairs(names)

if __name__=="__main__":
    main()

Answer 5

>>> from random import shuffle
>>> valid_names=["John", "Aaron", "Lydia", "Robin"]
>>> shuffle(valid_names)
>>> for pair in zip(valid_names, (valid_names*2)[1:]):
...     print(pair)
...
('Aaron', 'John')
('John', 'Lydia')
('Lydia', 'Robin')
('Robin', 'Aaron')

Answer 6

最简单的方法是将名称洗牌并将每个名称分配给下一个。它还保证两个人不会有彼此的名字。

>>> from random import shuffle
>>> names=['John', 'Aaron', 'Lydia', 'Robin']
>>> shuffle(names)
>>> dict(zip(names, names[-1:] + names))
{'Aaron': 'Lydia', 'John': 'Robin', 'Lydia': 'John', 'Robin': 'Aaron'}