小编ett*_*any的帖子

所以,我有这个代码:

url = 'http://google.com'
linkregex = re.compile('<a\s*href=[\'|"](.*?)[\'"].*?>')
m = urllib.request.urlopen(url)
msg = m.read()
links = linkregex.findall(msg)

但是然后python返回这个错误:

links = linkregex.findall(msg)
TypeError: can't use a string pattern on a bytes-like object

我做错了什么？

我有一个清单列表:

List = [-2,9,4,-6,7,0,1,-4]

对于列表中小于零(0)的数字,我想跳过这些数字并形成另一个列表.

例:-

List = [9,4,7,0,1]

这是我的一种怀疑,不确定如果我们能够实现.如果有可能实现,任何人都可以在这里发布.

我有一个3.5 go的大型csv文件,我想用pandas来阅读它.

这是我的代码:

import pandas as pd
tp = pd.read_csv('train_2011_2012_2013.csv', sep=';', iterator=True, chunksize=20000000, low_memory = False)
df = pd.concat(tp, ignore_index=True)

我收到此错误:

pandas/parser.pyx in pandas.parser.TextReader.read (pandas/parser.c:8771)()

pandas/parser.pyx in pandas.parser.TextReader._read_rows (pandas/parser.c:9731)()

pandas/parser.pyx in pandas.parser.TextReader._tokenize_rows (pandas/parser.c:9602)()

pandas/parser.pyx in pandas.parser.raise_parser_error (pandas/parser.c:23325)()

CParserError: Error tokenizing data. C error: out of 

我的公羊的容量是8 Go.

我有一个如下数据框:

  Index      Return
2008-11-21   0.153419
2008-11-24   0.037421
2008-11-25   0.077500

计算最后一行所有列的累积回报的最佳方法是什么？

以下是预期结果:

  Index      Return
2008-11-21   0.153419
2008-11-24   0.037421
2008-11-25   0.077500
Cumulative   0.289316

累积回报计算如下:

cumulative = (1 + return1) * (1 + return2) * (1 + return3) - 1 

在熊猫中执行此操作的最佳方法是什么？

我是Django的新手,我收到了标题中写的错误.我谷歌它,但我找不到任何东西.实际上我想要做的只是添加"评论"到我的博客网站.我添加我的代码如下.谢谢我的文件:

#form.py

from django import forms
from django.db import models
from blog.models import Comment


class EmailPostForm(forms.Form):
    name = forms.CharField(max_length=25)
    email = forms.EmailField()
    to = forms.EmailField()
    comments = forms.CharField(required=False, widget=forms.Textarea)
    email = forms.EmailField()
    to = forms.EmailField()
    comments = forms.CharField(required=False, widget=forms.Textarea)


class CommentForm(forms.ModelForm):
    class Meta:
        model = Comment
        fields = ('name', 'email', 'body ')

#models.py

from django.db import models
from django.utils import timezone
from django.contrib.auth.models import User
from django.core.urlresolvers import reverse

class Post(models.Model):
    STATUS_CHOICES = (
            ('draft', 'Draft'),
            ('published', 'Published'),
            )
    title …

如何在 Django 中捕获包含 yyyy-mm-dd 的 url。所以就像 www.mydomain.com/2011-02-12。我试过：

url(r'^(?P<date>\d{4}-\d{2}-\d{2})/$', views.index, name='index'), 

但是，服务器说找不到页面。

我是Python和Pandas的新手.我有一个熊猫数据框,每月的列从2000(2000-01)到2016(2016-06).

我想找到每三个月的平均值并将其分配给新的季度列(2000q1).我知道我可以做到以下几点:

df['2000q1'] = df[['2000-01', '2000-02', '2000-03']].mean(axis=1)
df['2000q2'] = df[['2000-04', '2000-05', '2000-06']].mean(axis=1)
    .
    .
    .
df['2016-02'] = df[['2016-04', '2016-05', '2016-06']].mean(axis=1)

但是,这非常乏味.如果有人帮我找到更好的方法,我感激不尽.

我试过这样但输出错误,请帮帮我

String inputString1 = "dfgh";// but not dFgH
String regex = "[a-z]";
boolean result;

Pattern pattern1 = Pattern.compile(regex);
Matcher matcher1 = pattern1.matcher(inputString1);
result = matcher1.matches();
System.out.println(result);

我有这个HTML代码:

<div class="first div">
    <div class="second">  
        <div class="title">Hi</div>
        <div class="test-icon" style="background-image: url(https://1.jpg);"></div>
    </div>
</div>
<div class="first div">
    <div class="second">  
        <div class="title">Hi</div>
        <div class="test-icon" style="background-image: url(https://12.jpg);"></div>
    </div>
</div>
<div class="first div">
    <div class="second">  
        <div class="title">Hi</div>
        <div class="test-icon" style="background-image: url(https://123.jpg);"></div>
    </div>
</div>
<div class="first div">
    <div class="second">  
        <div class="title">Hi</div>
        <div class="test-icon" style="background-image: url(https://good.jpg);"></div>
    </div>
</div>

如果背景图像不是,则可以删除所有带"first div"类的元素:url(https://good.jpg);？所以最终的回应是:

<div class="first div">
    <div class="second">  
        <div class="title">Hi</div>
        <div class="test-icon" style="background-image: url(https://good.jpg);"></div>
    </div>
</div>

我将不胜感激,谢谢!

我正在尝试从一组唯一值构建一个dict作为键和一个压缩的元组列表来提供项目.

set = ("a","b","c")

lst 1 =("a","a","b","b","c","d","d")
lst 2 =(1,2,3,3,4,5,6,)

zip = [("a",1),("a",2),("b",3),("b",3),("c",4),("d",5)("d",6)

dct = {"a":1,2 "b":3,3 "c":4 "d":5,6}

但我得到:

dct = {"a":1,"b":3,"c":4,"d":5}

这是我目前的代码:

#make two lists 

rtList = ["EVT","EVT","EVT","EVT","EVT","EVT","EVT","HIL"]
raList = ["C64G","C64R","C64O","C32G","C96G","C96R","C96O","RA96O"]

# make a set of unique codes in the first list

routes = set()
for r in rtList:
    routes.add(r)

#zip the lists

RtRaList = zip(rtList,raList)
#print RtRaList

# make a dictionary with list one as the keys and list two as the values

SrvCodeDct = {}

for …