小编jam*_*lak的帖子

所以当我尝试从我的类中获取列表的len()时出现此错误:TypeError:当我尝试时,类型为'Stuff'的对象没有len():

>>> s = Stuff()
>>> len(s)
error instead of showing 0 like:
>>> l = []
>>> len(l)
0

码:

class Stuff():
    def __init__(self):
        self.lst = []

每个人我只是想知道如何计算str中ch的重复,我写的函数是:

def repeat(s):
    '''
    '''
    count = 0 
    L = []

    for i in range(len(s)):
        count = 0
        if s[i] in s: 
         count = count + 1
        else:
             count = count

    L.append(count)
    return L

但是,如果字母相同,列表将重复.是否有更好的方法来计算str中的字符数量？我在考虑使用字典？

此程序计算2的平方根,但它不会将数字显示为我想要的1000位精度,而是只显示50位数.

这是程序

import math 
a=format(2.0**0.5, '.1000')   
print a

我想知道如何编写程序来显示更多数字.

>>> a = (1,2)
>>> a += (3,4)
>>> a
(1, 2, 3, 4)
>>> 

并列出:

>>> b = [1,2]
>>> b += [3,4]
>>> b
[1, 2, 3, 4]
>>> 

由于元组是不可变的并且列表是可变的,我们如何解释这种行为？

我有这样的2个列表:

list1=['A','B'] 
list2=['A','C','D']

我想在两个列表中进行比较,找出缺少的,额外的和条目没有变化,我这样访问它:

set1=set(list1)
set2=set(list2)

MissingName=set1.difference(set2)
AdditionalName=set2.difference(set1)

这给了我遗漏和额外的条目,我怎么能找到无变化,哪个应该是A ??

我正在尝试循环遍历python中的列表,并将列表中的每个元素与列表中的下一个15个元素进行比较.我想要类似下面的代码,但我不知道如何在python中说出来.另外,我想在不进行任何额外输入的情况下这样做.

for i in S:
    for j in S[i: '''up 15 elements after i or however much is left if less than 15''']
        #do some comparisons

所以基本上,对于内部循环,我只想从项目i迭代到项目i + 15(或者剩下的任何东西),而不需要导入迭代器等.

谢谢!

在Python 2.5中,我使用了代码:

    yn=raw_input('''This program is for solving linear and quadratic equations.
    Is this what you want to do?
    ''')
    if not 'y' in yn or 'Y' in yn:
        print 'Program now quitting.'

Program now quitting当程序关闭时显示,当yn不包含'y'或'Y'时应该显示,但是当yn确实包含'Y'时,它也显示在程序的其余部分开始之前.我认为错误是通过使用"或",造成"不"在一起,因为它并没有改变,当我改变if not 'y' in yn or 'Y' in yn:到if not 'y' in yn or not 'Y' in yn:,但消失时,我把它改成

    if not 'y' in yn:
        if not 'Y' in yn:

这是正确的吗？如果有的话,我可以做些什么来解决问题并将这部分保持在同一条线上？

我试图找到一个点[x1,y1]和以[c1,c2]为中心的半径为2的圆上最近点之间的距离？我为此设计了以下等式

sqrt((p[0] - c[0]) ** 2 + (p[1] - c[1]) ** 2)

但答案却回答了一个似乎错误的答案......

我究竟做错了什么？

使用.read()读取文件，我将如何同时对两个物体分开？我试图同时拆分逗号，"\n"但是当我首先拆分逗号时，它会将我的字符串变成一个列表，在其中我无法再次拆分。

这是我试图拆分的字符串：

'States, Total Score, Critical Reading, Mathematics, Writing, Participation (%)\nWashington,1564,524,532,508,41.2000\nNewHampshire,1554,520,524,510,64.0000\nMassachusetts,1547,512,526,509,72.1000\nOregon,1546,523,524,499,37.1000\nVermont,1546,519,512,506,64.0000\nArizona,1544,519,525,500,22.4000\nConnecticut,1536,509,514,513,71.2000\nAlaska,1524,518,515,491,32.7000\nVirginia,1521,512,512,497,56.0000\nCalifornia,1517,501,516,500,37.5000\nNewJersey,1506,495,514,497,69.0000\nMaryland,1502,501,506,495,56.7000\nNorthCarolina,1485,497,511,477,45.5000\nRhodeIsland,1477,494,495,488,60.8000\nIndiana,1476,494,505,477,52.0000\nFlorida,1473,496,498,479,44.7000\nPennsylvania,1473,492,501,480,62.3000\nNevada,1470,496,501,473,25.9000\nDelaware,1469,493,495,481,59.2000\nTexas,1462,484,505,473,41.5000\nNewYork,1461,484,499,478,59.6000\nHawaii,1458,483,505,470,47.1000\nGeorgia,1453,488,490,475,46.5000\nSouthCarolina,1447,484,495,468,40.7000\nMaine,1389,468,467,454,87.1000\nIowa,1798,603,613,582,2.7000\nMinnesota,1781,594,607,580,6.0000\nWisconsin,1778,595,604,579,3.8000\nMissouri,1768,593,595,580,3.6000\nMichigan,1766,585,605,576,3.8000\nSouthDakota,1766,592,603,571,2.0000\nIllinois,1762,585,600,577,4.6700\nKansas,1752,590,595,567,4.7000\nNebraska,1746,585,593,568,3.9000\nNorthDakota,1733,580,594,559,3.4000\nKentucky,1713,575,575,563,5.0000\nTennessee,1712,576,571,565,6.4000\nColorado,1695,568,572,555,14.1000\nArkansas,1684,566,566,552,3.5000\nOklahoma,1684,569,568,547,3.8000\nWyoming,1683,570,567,546,3.6000\nUtah,1674,568,559,547,4.5000\nMississippi,1666,566,548,552,2.2000\nLouisiana,1652,555,550,547,4.0000\nAlabama,1650,556,550,544,5.4000\nNewMexico,1636,553,549,534,7.1000\nOhio,1609,538,548,522,17.2000\nIdaho,1601,543,541,517,14.6000\nMontana,1593,538,538,517,20.0000\nWest Virginia,1522,515,507,500,13.2000\n'

将参数分成行的惯例是什么？是否有PEP对此进行制裁？我没有在PEP8中找到它.

file_like = f(self,
              path,
              mode=mode,
              buffering=buffering,
              encoding=encoding,
              errors=errors,
              newline=newline,
              line_buffering=line_buffering,
              **kwargs)