小编Mar*_*ers的帖子

条件None == None是真还是假？

我有 2 个熊猫数据框：

import pandas as pd

df1 = pd.DataFrame({'id':[1,2,3,4,5], 'value':[None,20,None,40,50]})
df2 = pd.DataFrame({'index':[1,2,3], 'value':[None,20,None]})

In [42]: df1

Out[42]:    id  value
         0   1    NaN
         1   2   20.0
         2   3    NaN
         3   4   40.0
         4   5   50.0

In [43]: df2

Out[43]:    index  value
         0      1    NaN
         1      2   20.0
         2      3    NaN

当我执行合并操作时，它看起来None == None是 True：

In [37]: df3 = df1.merge(df2, on='value', how='inner')
In [38]: df3
Out[38]:    id  value  index
         0   1    NaN      1 …

我不明白为什么要运行Flask应用程序

(venv) $ export FLASK_APP=microblog.py
(venv) $ flask run

但是如果我想运行Django应用程序，我只会

(venv) $ python manage.py runserver

没有export DJANGO_APP=microblog.py
为什么？为什么在第一种情况下我需要导出应用程序，但在第二种情况下我不需要？

我希望有一个7/10的机会成功完成一个动作,比如爬墙.我无法使用random.randint获得一组整数,所以我不得不重新键入10次,如下所示:

import random

print('You try to climb the wall.')

climbResult = random.randint(1,10)


elif climbResult == (1):
    print('You were successful.')

elif climbResult == (2):
    print('You were successful.')

elif climbResult == (3):
    print('You were successful.')

elif climbResult == (4):
    print('You were successful.')

elif climbResult == (5):
    print('You were successful.')

elif climbResult == (6):
    print('You were successful.')

elif climbResult == (7):
    print('You were successful.')

elif climbResult == (8):
    print('You were unsuccessful.')

elif climbResult == (9):
    print('You were unsuccessful.')

elif climbResult == (10):
    print('You …

你能看看以下内容:

{'2013-01-02': {'Volume': '4202600000', 'Adj Close': '1462.42', 'High': '1462.43', 'Low': '1426.19', 'Close': '1462.42', 'Open': '1426.19'}}

这是什么？不是一个肯定的清单,我不能这样做:

print [0]

例如..

任何帮助,将不胜感激!

我在codeblocks上测试了这段代码并且工作正常但是当我在Ideone上运行时,我在OJ上遇到运行时错误和SIGSEGV错误.我在网上看到SIGSEGV错误是由于内存访问受限造成的.但如果它这样做,为什么不是代码块抱怨......

#include <bits/stdc++.h>
using namespace std;

int main()
{
    ios_base::sync_with_stdio(false);
    int t, i, j, k, x, temp;
    int num1[10000], num2[10000];
    char c;
    bool f = true;
    cin >> t;

    while (t--)
    {
        for (i = 0; i < 10000; i++)
        {
            num1[i] = 0;
            num2[i] = 0;
        }
        i = 0;
        j = 0;
        while (!isspace(c = getchar()))
            num1[i++] = c - '0';
        while (!isspace(c = getchar()))
            num2[j++] = c - '0';
        x = i > j ? i : …

我的代码有一些问题,看看main()部分,我输入选择2,它调用open_file.但它需要1作为别的而不是IF ...而当我用1 agin写它只是在屏幕上打印1,我做错了什么？Python版本3.4.2.

import sys
name=input("What is your name? : ")
print ("Welcome " + name)


def main():
    print ("What do you want to do")
    print ("1) Open A File")
    print ("2) open Web Browser")
    print ("3) Exit OS")

    main_choice=input()

    if main_choice == 1:
        open_file()
    elif main_choice == 2:
        web_browser()
    elif main_choice == 3:
        exit_os()
    else:
        unknown_number()


def unknown_number():
    print ("The choice you made does not exist, please choose a valid option")
    main


def open_file():
    print ("What do you want …

我有一个列表字典,summary:

summary = {
    'Raonic': [2, 0, 11, 122, 16, 139],
    'Halep': [2, 2, 10, 75, 6, 60],
    'Kerber': [2, 0, 7, 68, 7, 71],
    'Wawrinka': [1, 2, 14, 133, 13, 128],
    'Djokovic': [2, 2, 10, 75, 8, 125],
}

我希望打印到屏幕(标准输出)按排名降序排列的摘要,其中排名按照该顺序中的条件1-6(比较项目1(列表的)),如果相等比较项目2 (列表中的),如果它们相等,则比较项目3(列表中的).该比较按降序继续到列表的项目4 .

但是,列表的第5项和第6项的比较必须按升序进行.

输出:

Halep 2 2 10 75 6 60
Djokovic 2 2 10 75 8 125
Raonic 2 0 11 122 16 139
Kerber 2 0 7 68 7 71
Wawrinka …

我想检查我的值是否是带有点或逗号的浮点数,但是isdigit()返回带有点的false.我想知道为什么以及如何通过它.

> value = "0.0"
> print value.isdigit():
>>> False

我的代码是:

if "." in value and value.isdigit()
    print "ok"

请看以下内容:

def update_page_info(url):
    # fetch_page -> parse_page -> store_page
    chain = fetch_page.s(url) | parse_page.s() | store_page_info.s(url)
    chain()

@app.task()
def fetch_page(url):
    return myhttplib.get(url)

@app.task()
def parse_page(page):
    return myparser.parse_document(page)

@app.task(ignore_result=True)
def store_page_info(info, url):
    PageInfo.objects.create(url=url, info=info)

在速度和代码优化方面有什么更好的选择？执行计算，将结果参考存储在本地，然后像这样返回参考：

def aplusb(a,b):
    result = a + b
    return result

或将计算直接返回给调用者的结果返回：

def aplusb(a,b):
    return a + b