小编Bio*_*eek的帖子

假设我们有一个翻译莫尔斯符号的函数:

• . - > -.
• - - > ...-

如果我们两次应用此函数,我们得到例如:

.- > -.- >...--.

给定输入字符串和重复次数,想知道最终字符串的长度.(佛兰芒编程竞赛 VPW中的问题1 ,取自这些在Haskell中提供解决方案的幻灯片).

对于给定的输入文件

4
. 4
.- 2
-- 2 
--... 50

我们期待解决方案

44
16
20
34028664377246354505728

由于我不知道Haskell,这是我在Python中提出的递归解决方案:

def encode(msg, repetition, morse={'.': '-.', '-': '...-'}):
    if isinstance(repetition, str):
        repetition = eval(repetition)
    while repetition > 0:
        newmsg = ''.join(morse[c] for c in msg)
        return encode(newmsg, repetition-1)
    return len(msg)


def problem1(fn):
    with open(fn) as f:
        f.next()
        for line in …

我有一组n(~1000000)字符串(DNA序列)存储在列表trans中.我必须找到列表中所有序列的最小汉明距离.我实施了一个天真的暴力算法,它运行了一天多,还没有给出解决方案.我的代码是

dmin=len(trans[0])
for i in xrange(len(trans)):
    for j in xrange(i+1,len(trans)):
            dist=hamdist(trans[i][:-1], trans[j][:-1])
            if dist < dmin:
                    dmin = dist

有没有更有效的方法来做到这一点？Hamdist是我写的一个函数,用于查找汉明距离.它是

def hamdist(str1, str2):
    diffs = 0
    if len(str1) != len(str2):
        return max(len(str1),len(str2))
    for ch1, ch2 in zip(str1, str2):
        if ch1 != ch2:
          diffs += 1
    return diffs

我明白,我应该使用os.urandom()或SystemRandomPython中的"安全"的伪随机数.

但是Python如何在逻辑上生成这些随机数？

还有一种方法可以在Python中生成一个比其他更"随机"的数字吗？

我有一个简单的 if-else 语句

if(this.props.params.id){
    //do something
}

现在，这在浏览器中运行良好。如果参数中有 id，则进入 if 子句；如果没有 id，则不进入 if 子句。

现在，在用 jest 编写测试时，当未定义 id 时，它会抛出错误：“无法读取未定义的属性'id'”为什么会发生这种情况，不应该将其视为 false 吗？它在浏览器中运行良好，只是在测试时会抛出错误。

我numpy从http://www.lfd.uci.edu/~gohlke/pythonlibs/#numpy下载了一个预编译的二进制文件,并尝试numpy在Windows 7上升级我当前的安装程序

pip install --upgrade "numpy-1.10.4 vanilla-cp27-none-win32.whl"

我收到以下错误:

C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\Scripts\pip-script.py run on 04/01/16 13:20:05
Unpacking c:\users\jeroen\downloads\numpy-1.10.4 vanilla-cp27-none-win_amd64.whl
Installing collected packages: numpy
Cleaning up...
Exception:
Traceback (most recent call last):
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\basecommand.py", line 122, in main
    status = self.run(options, args)
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\commands\install.py", line 283, in run
    requirement_set.install(install_options, global_options, root=options.root_path)
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\req.py", line 1435, in install
    requirement.install(install_options, global_options, *args, **kwargs)
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\req.py", line 671, in install
    self.move_wheel_files(self.source_dir, root=root)
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\req.py", line 901, in move_wheel_files
    pycompile=self.pycompile,
  File "C:\Users\Jeroen\AppData\Local\Enthought\Canopy\User\lib\site-packages\pip\wheel.py", …

我有一个带有输入字段的Bootstrap 4表单,称为runname.我想在输入字段上执行以下验证:

• runname 不能为空
• runname 不能包含空格
• runname以前不能使用

我已经有一个表单的代码,如果输入字段为空,则使用自定义Bootstrap样式给出错误:

// JavaScript for disabling form submissions if there are invalid fields
(function() {
  'use strict';
  window.addEventListener('load', function() {
    // Fetch all the forms we want to apply custom Bootstrap validation styles to
    var forms = document.getElementsByClassName('needs-validation');
    // Loop over them and prevent submission
    var validation = Array.prototype.filter.call(forms, function(form) {
      form.addEventListener('submit', function(event) {
        if (form.checkValidity() === false) {
          event.preventDefault();
          event.stopPropagation();
        }
        form.classList.add('was-validated');
      }, false);
    });
  }, false);
})(); …

我向CareerBuilder API发送GET请求:

import requests

url = "http://api.careerbuilder.com/v1/jobsearch"
payload = {'DeveloperKey': 'MY_DEVLOPER_KEY',
           'JobTitle': 'Biologist'}
r = requests.get(url, params=payload)
xml = r.text

并获得一个看起来像这样的XML .但是,我无法解析它.

使用其中之一 lxml

>>> from lxml import etree
>>> print etree.fromstring(xml)

Traceback (most recent call last):
  File "<pyshell#4>", line 1, in <module>
    print etree.fromstring(xml)
  File "lxml.etree.pyx", line 2992, in lxml.etree.fromstring (src\lxml\lxml.etree.c:62311)
  File "parser.pxi", line 1585, in lxml.etree._parseMemoryDocument (src\lxml\lxml.etree.c:91625)
ValueError: Unicode strings with encoding declaration are not supported.

要么 ElementTree:

Traceback (most recent call last):
  File …

我有一个像这样的列表列表

matches = [[['rootrank', 'Root'], ['domain', 'Bacteria'], ['phylum', 'Firmicutes'], ['class', 'Clostridia'], ['order', 'Clostridiales'], ['family', 'Lachnospiraceae'], ['genus', 'Lachnospira']], 
           [['rootrank', 'Root'], ['domain', 'Bacteria'], ['phylum', '"Proteobacteria"'], ['class', 'Gammaproteobacteria'], ['order', '"Vibrionales"'], ['family', 'Vibrionaceae'], ['genus', 'Catenococcus']], 
           [['rootrank', 'Root'], ['domain', 'Archaea'], ['phylum', '"Euryarchaeota"'], ['class', '"Methanomicrobia"'], ['order', 'Methanomicrobiales'], ['family', 'Methanomicrobiaceae'], ['genus', 'Methanoplanus']]]

我想从它们构建系统发育树。我写了一个像这样的节点类（部分基于此代码）：

class Node(object):
    """Generic n-ary tree node object
    Children are additive; no provision for deleting them."""

    def __init__(self, parent, category=None, name=None):
        self.parent = parent
        self.category = category
        self.name = name
        self.childList = [] …

我有一个列表,可以包含Nones和datetime对象.我需要在连续datetime对象的子列表中拆分它,并需要datetime在原始列表中记录该子列表的第一个对象的索引.

例如,我需要能够转向

original = [None, datetime(2013, 6, 4), datetime(2014, 5, 12), None, None, datetime(2012, 5, 18), None]

成:

(1, [datetime.datetime(2013, 6, 4, 0, 0), datetime.datetime(2014, 5, 12, 0, 0)])
(5, [datetime.datetime(2012, 5, 18, 0, 0)])

我尝试了两种方法.一个使用find:

binary = ''.join('1' if d else '0' for d in original)
end = 0
start = binary.find('1', end)
while start > -1:
    end = binary.find('0', start)
    if end < 0:
        end = len(binary)
    dates …

我有一个自定义对象列表,我想从中删除重复项.通常情况下,你会被定义都做到这一点__eq__,并__hash__为您的对象,然后取set对象的列表中.我已经定义了__eq__,但是我无法找到一种好的方法来实现__hash__它为相等的对象返回相同的值.

更具体地说,我有一个派生自ete3工具包中的Tree类的类.如果Robinson-Foulds距离为零,我已将两个对象定义为相等.

from ete3 import Tree

class MyTree(Tree):

    def __init__(self, *args, **kwargs):
        super(MyTree, self).__init__(*args, **kwargs)

    def __eq__(self, other):
        rf = self.robinson_foulds(other, unrooted_trees=True)
        return not bool(rf[0])

newicks = ['((D, C), (A, B),(E));',
           '((D, B), (A, C),(E));',
           '((D, A), (B, C),(E));',
           '((C, D), (A, B),(E));',
           '((C, B), (A, D),(E));',
           '((C, A), (B, D),(E));',
           '((B, D), (A, C),(E));',
           '((B, C), (A, D),(E));',
           '((B, A), (C, D),(E));',
           '((A, …