Iterate over a dictionary in Python

This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers You iterate through a dictionary just like an array, but instead of giving you the values in the dictionary it gives you the keys. > my_dict = {"a" : 4, "b" : 7, "c" : 8} > for i in my_dict: ... print i a b c You can then access the data in the dictionary like you normally woul

在Python中迭代字典

这个问题在这里已经有了答案: 使用'for'循环遍历字典12个答案 您可以像数组一样遍历字典,但不会向字典中的值提供键。 > my_dict = {"a" : 4, "b" : 7, "c" : 8} > for i in my_dict: ... print i a b c 然后,您可以像往常一样访问字典中的数据。 (这是用方括号来实现的,所以my_dict["a"]会给出4) 字典有能力遍历自己打印每个项目。 要按照您的要求简单打印每个项目的详细信息,只需使

How to access my dictionary

This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers Discover is a key of your dictionary. You can access the keys collection in your dictionary using keys() method. Then use each key to access the associated element. for key in account.keys(): print(key) print(account[key]) Since Discover is the only key you could do print acc

如何访问我的字典

这个问题在这里已经有了答案: 使用'for'循环遍历字典12个答案 Discover是你的词典的关键。 您可以使用keys()方法访问字典中的密钥集合。 然后使用每个键访问关联的元素。 for key in account.keys(): print(key) print(account[key]) 由于Discover是您可以print account.keys()[0]的唯一键。 但是,如果你有更多的键,这将不会每次都工作,因为字典是任意排序的(换句话说,没有秩序感),你需要迭

Use more than 1 iterable in a python for loop

This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers Get the cartesian product of a series of lists? 10 answers for files,script in TEMPLATE_FILE.items(): print(files,scripts) is the construction you're looking for. (in python 2 there's an iteritems which is removed in python 3 so for small dictionaries items is OK and po

在python for循环中使用多于一个迭代

这个问题在这里已经有了答案: 使用'for'循环遍历字典12个答案 获取一系列列表的笛卡尔积? 10个答案 for files,script in TEMPLATE_FILE.items(): print(files,scripts) 是你正在寻找的建筑。 (在python 2中有一个iteritems在python 3中被删除,所以对于小字典items是可以的并且是可移植的) 当然你可以这样做: for files in TEMPLATE_FILE: scripts = TEMPLATE_FILE[files] 但这并不像你在每

Join dictionary item, keys

This question already has an answer here: Iterating over dictionaries using 'for' loops 12 answers Python dictionaries are unordered (or rather, their order is arbitrary), and when you iterate on them, only the keys are returned: >>> d = {'0':0, '1':1, '2':2, '3':3, '4':4} >>> print(d) {'4': 4, '1': 1, '0': 0, '3': 3, '2': 2} If you need both keys and values, use i

加入词典项目,键

这个问题在这里已经有了答案: 使用'for'循环遍历字典12个答案 Python字典是无序的(或者说,它们的顺序是任意的),并且当你迭代它们时,只返回键值: >>> d = {'0':0, '1':1, '2':2, '3':3, '4':4} >>> print(d) {'4': 4, '1': 1, '0': 0, '3': 3, '2': 2} 如果您需要键和值,请使用iDict.items() 。 如果您需要订购,请使用collections.OrderedDict 。 对字典的迭代只能产生密钥: >&

Why does the **kwargs mapping compare equal with a differently ordered OrderedDict?

According to PEP 468: Starting in version 3.6 Python will preserve the order of keyword arguments as passed to a function. To accomplish this the collected kwargs will now be an ordered mapping . Note that this does not necessarily mean OrderedDict . In that case, why does this ordered mapping fail to respect equality comparison with Python's canonical ordered mapping type, the collecti

为什么** kwargs映射比较与OrderedDict有所不同?

根据PEP 468: 从版本3.6开始,Python将保留传递给函数的关键字参数的顺序。 为了实现这一点,收集的kwargs现在将成为一个有序映射 。 请注意,这并不一定意味着OrderedDict 。 在那种情况下,为什么这个有序映射无法尊重与Python的规范有序映射类型( collections.OrderedDict平等比较: >>> from collections import OrderedDict >>> data = OrderedDict(zip('xy', 'xy')) >>> def foo(**kwa

How to set environment variables in Python

I need to set some environment variables in the python script and I want all the other scripts that are called from python (shell scripts) which will be child process to see the environment variables set. The value is a number. If I do os.environ["DEBUSSY"] = 1 , it complains saying that 1 has to be string. I also want to know how to read the environment variables in python (in the

如何在Python中设置环境变量

我需要在python脚本中设置一些环境变量,并且需要从python(shell脚本)调用的所有其他脚本,这些脚本将是子进程,以查看环境变量集。 该值是一个数字。 如果我做os.environ["DEBUSSY"] = 1 ,它会抱怨说1必须是字符串。 我也想知道如何在Python中设置它的环境变量(在脚本的后面部分)。 环境变量必须是字符串,所以使用 os.environ["DEBUSSY"] = "1" 将变量DEBUSSY设置为字符串1 。 稍后要访问此变量,只需

MongoDB InvalidDocument: Cannot encode object

I am using scrapy to scrap blogs and then store the data in mongodb. At first i got the InvalidDocument Exception. So obvious to me is that the data is not in the right encoding. So before persisting the object, in my MongoPipeline i check if the document is in 'utf-8 strict', and only then i try to persist the object to mongodb. BUT Still i get InvalidDocument Exceptions, now that is

MongoDB InvalidDocument:无法编码对象

我正在使用scrapy来取消博客,然后将数据存储在mongodb中。 起初我得到了InvalidDocument异常。 对我来说非常明显的是,数据不是正确的编码。 所以在持久化对象之前,在我的MongoPipeline中,我检查文档是否在'utf-8 strict'中,然后我尝试将对象保存到mongodb中。 但仍然我得到InvalidDocument异常,现在很烦人。 这是我的代码,我的MongoPipeline对象持久对象mongodb # -*- coding: utf-8 -*- # Define your it

OpenCV with AWS Lambda

I wrote a Lambda function (Python 2.7) that uses OpenCV. I need to import CV2 and I'm having trouble providing the library in my .zip. Has anyone used OpenCV with Lambda? How can I provide the CV2 module. You must copy the OpenCV inside the zip that you send to AWS Lambda This method helps to deploy with OpenCV https://github.com/aeddi/aws-lambda-python-opencv/blob/master/build.sh Yo

使用AWS Lambda的OpenCV

我写了一个使用OpenCV的Lambda函数(Python 2.7)。 我需要导入CV2,并且无法在我的.zip中提供库。 有没有人用Lambda使用OpenCV? 我如何提供CV2模块。 您必须将OpenCV复制到发送给AWS Lambda的zip中 此方法有助于使用OpenCV进行部署https://github.com/aeddi/aws-lambda-python-opencv/blob/master/build.sh 您需要在Amazon Linux环境中构建您的代码。 请参阅我的回答:https://stackoverflow.com/a/39649170/99692

Delimit a specific column and add them as columns in CSV (Python3, CSV)

I have a csv file that has several columns that I first delimit by colon (;). However, ONE column is delimited by a pipe | and I would like to delimit this column and create new columns. Input: Column 1 Column 2 Column 3 1 2 3|4|5 6 7 6|7|8 10 11 12|13|14 Desired Output: Column 1 Column 2 ID Age He

划定特定列并将它们添加为CSV(Python3,CSV)列

我有一个csv文件,它有几列,我首先用冒号(;)分隔。 但是,ONE列由管道分隔 我想分隔这个列并创建新的列。 输入: Column 1 Column 2 Column 3 1 2 3|4|5 6 7 6|7|8 10 11 12|13|14 期望的输出: Column 1 Column 2 ID Age Height 1 2 3 4 5 6 7 6 7 8

How do I use python

I recently installed python-WikEdDiff package to my system. I understand it is a python extension of the original JavaScript WikEdDiff tool. I tried to use it but I couldn't find any documentation for it. I am stuck at using WikEdDiff.diff() . I wish to use the other functions of this class, such as getFragments() and others, but on checking, it shows the following error: Traceback (mos

我如何使用python

我最近在我的系统中安装了python-WikEdDiff软件包。 我知道它是原始JavaScript WikEdDiff工具的一个python扩展。 我试图使用它,但我找不到任何文档。 我坚持使用WikEdDiff.diff() 。 我希望使用这个类的其他函数,比如getFragments()和其他函数,但是在检查时会显示以下错误: Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/local/lib/python3.4/dist-packages/Wik