Python 实用宝典

Question 1

I am running unit tests on a CI server using py.test. Tests use external resources fetched over network. Sometimes test runner takes too long, causing test runner to be aborted. I cannot repeat the issues locally.

Is there a way to make py.test print out execution times of (slow) test, so pinning down problematic tests become easier?

Question 2

I’m not sure this will solve your problem, but you can pass --durations=N to print the slowest N tests after the test suite finishes.

Use --durations=0 to print all.

Question 3

You can pass the number with --durations

pytest --durations=0 — Show all times for tests and setup and teardown

pytest --durations=1 — Just show me the slowest

pytest --durations=50 — Slowest 50, with times, … etc

Take refer in: https://medium.com/@brianokken/pytest-durations-0-show-all-times-for-tests-and-setup-and-teardown-848dccac85db

Or: https://docs.pytest.org/en/latest/usage.html#profiling-test-execution-duration

Question 4

How to concatenate strings in python?

For example:

Section = 'C_type'

Concatenate it with Sec_ to form the string:

Sec_C_type

Question 5

The easiest way would be

Section = 'Sec_' + Section

But for efficiency, see: https://waymoot.org/home/python_string/

Question 6

you can also do this:

section = "C_type"
new_section = "Sec_%s" % section

This allows you not only append, but also insert wherever in the string:

section = "C_type"
new_section = "Sec_%s_blah" % section

Question 7

Just a comment, as someone may find it useful – you can concatenate more than one string in one go:

>>> a='rabbit'
>>> b='fox'
>>> print '%s and %s' %(a,b)
rabbit and fox

Question 8

More efficient ways of concatenating strings are:

join():

Very efficent, but a bit hard to read.

>>> Section = 'C_type'  
>>> new_str = ''.join(['Sec_', Section]) # inserting a list of strings 
>>> print new_str 
>>> 'Sec_C_type'

String formatting:

Easy to read and in most cases faster than ‘+’ concatenating

>>> Section = 'C_type'
>>> print 'Sec_%s' % Section
>>> 'Sec_C_type'

Question 9

Use + for string concatenation as:

section = 'C_type'
new_section = 'Sec_' + section

Question 10

To concatenate strings in python you use the “+” sign

ref: http://www.gidnetwork.com/b-40.html

Question 11

For cases of appending to end of existing string:

string = "Sec_"
string += "C_type"
print(string)

results in

Sec_C_type

Question 12

If you have a collection of methods in a file, is there a way to include those files in another file, but call them without any prefix (i.e. file prefix)?

So if I have:

[Math.py]
def Calculate ( num )

How do I call it like this:

[Tool.py]
using Math.py

for i in range ( 5 ) :
    Calculate ( i )

Question 13

You will need to import the other file as a module like this:

import Math

If you don’t want to prefix your Calculate function with the module name then do this:

from Math import Calculate

If you want to import all members of a module then do this:

from Math import *

Edit: Here is a good chapter from Dive Into Python that goes a bit more in depth on this topic.

Question 14

Just write the “include” command :

import os

def include(filename):
    if os.path.exists(filename): 
        execfile(filename)


include('myfile.py')

@Deleet :

@bfieck remark is correct, for python 2 and 3 compatibility, you need either :

Python 2 and 3: alternative 1

from past.builtins import execfile

execfile('myfile.py')

Python 2 and 3: alternative 2

exec(compile(open('myfile.py').read()))

Question 15

If you use:

import Math

then that will allow you to use Math’s functions, but you must do Math.Calculate, so that is obviously what you don’t want.

If you want to import a module’s functions without having to prefix them, you must explicitly name them, like:

from Math import Calculate, Add, Subtract

Now, you can reference Calculate, Add, and Subtract just by their names. If you wanted to import ALL functions from Math, do:

from Math import *

However, you should be very careful when doing this with modules whose contents you are unsure of. If you import two modules who contain definitions for the same function name, one function will overwrite the other, with you none the wiser.

Question 16

I’ve found the python inspect module to be very useful

For example with teststuff.py

import inspect

def dostuff():
    return __name__

DOSTUFF_SOURCE = inspect.getsource(dostuff)

if __name__ == "__main__":

    dostuff()

And from the another script or the python console

import teststuff

exec(DOSTUFF_SOURCE)

dostuff()

And now dostuff should be in the local scope and dostuff() will return the console or scripts _name_ whereas executing test.dostuff() will return the python modules name.

Question 17

What is the raw_input function? Is it a user interface? When do we use it?

Question 18

It presents a prompt to the user (the optional arg of raw_input([arg])), gets input from the user and returns the data input by the user in a string. See the docs for raw_input().

Example:

name = raw_input("What is your name? ")
print "Hello, %s." % name

This differs from input() in that the latter tries to interpret the input given by the user; it is usually best to avoid input() and to stick with raw_input() and custom parsing/conversion code.

Note: This is for Python 2.x

Question 19

raw_input() was renamed to input() in Python 3.

From http://docs.python.org/dev/py3k/whatsnew/3.0.html

Question 20

The “input” function converts the input you enter as if it were python code. “raw_input” doesn’t convert the input and takes the input as it is given. Its advisable to use raw_input for everything. Usage:

>>a = raw_input()
>>5
>>a
>>'5'

Question 21

raw_input is a form of input that takes the argument in the form of a string whereas the input function takes the value depending upon your input. Say, a=input(5) returns a as an integer with value 5 whereas a=raw_input(5) returns a as a string of “5”

Question 22

Another example method, to mix the prompt using print, if you need to make your code simpler.

Format:-

x = raw_input () — This will return the user input as a string

x= int(raw_input()) — Gets the input number as a string from raw_input() and then converts it to an integer using int().

print '\nWhat\'s your name ?', 
name = raw_input('--> ')
print '\nHow old are you, %s?' % name,
age = int(raw_input())
print '\nHow tall are you (in cms), %s?' % name,
height = int(raw_input())
print '\nHow much do you weigh (in kgs), %s?' % name,
weight = int(raw_input())

print '\nSo, %s is %d years old, %d cms tall and weighs %d kgs.\n' %(
name, age, height, weight)

Question 23

If I let raw_input like that, no Josh or anything else. It’s a variable,I think,but I don’t understand her roll :-(

The raw_input function prompts you for input and returns that as a string. This certainly worked for me. You don’t need idle. Just open a “DOS prompt” and run the program.

This is what it looked like for me:

C:\temp>type test.py
print "Halt!"
s = raw_input("Who Goes there? ")
print "You may pass,", s

C:\temp>python test.py
Halt!
Who Goes there? Magnus
You may pass, Magnus

I types my name and pressed [Enter] after the program had printed “Who Goes there?”

Question 24

Comprehensions are having some unexpected interactions with scoping. Is this the expected behavior?

I’ve got a method:

def leave_room(self, uid):
  u = self.user_by_id(uid)
  r = self.rooms[u.rid]

  other_uids = [ouid for ouid in r.users_by_id.keys() if ouid != u.uid]
  other_us = [self.user_by_id(uid) for uid in other_uids]

  r.remove_user(uid) # OOPS! uid has been re-bound by the list comprehension above

  # Interestingly, it's rebound to the last uid in the list, so the error only shows
  # up when len > 1

At the risk of whining, this is a brutal source of errors. As I write new code, I just occasionally find very weird errors due to rebinding — even now that I know it’s a problem. I need to make a rule like “always preface temp vars in list comprehensions with underscore”, but even that’s not fool-proof.

The fact that there’s this random time-bomb waiting kind of negates all the nice “ease of use” of list comprehensions.

Question 25

List comprehensions leak the loop control variable in Python 2 but not in Python 3. Here’s Guido van Rossum (creator of Python) explaining the history behind this:

We also made another change in Python 3, to improve equivalence between list comprehensions and generator expressions. In Python 2, the list comprehension “leaks” the loop control variable into the surrounding scope:
x = 'before'
a = [x for x in 1, 2, 3]
print x # this prints '3', not 'before'
This was an artifact of the original implementation of list comprehensions; it was one of Python’s “dirty little secrets” for years. It started out as an intentional compromise to make list comprehensions blindingly fast, and while it was not a common pitfall for beginners, it definitely stung people occasionally. For generator expressions we could not do this. Generator expressions are implemented using generators, whose execution requires a separate execution frame. Thus, generator expressions (especially if they iterate over a short sequence) were less efficient than list comprehensions.

However, in Python 3, we decided to fix the “dirty little secret” of list comprehensions by using the same implementation strategy as for generator expressions. Thus, in Python 3, the above example (after modification to use print(x) :-) will print ‘before’, proving that the ‘x’ in the list comprehension temporarily shadows but does not override the ‘x’ in the surrounding scope.

Question 26

Yes, list comprehensions “leak” their variable in Python 2.x, just like for loops.

In retrospect, this was recognized to be a mistake, and it was avoided with generator expressions. EDIT: As Matt B. notes it was also avoided when set and dictionary comprehension syntaxes were backported from Python 3.

List comprehensions’ behavior had to be left as it is in Python 2, but it’s fully fixed in Python 3.

This means that in all of:

list(x for x in a if x>32)
set(x//4 for x in a if x>32)         # just another generator exp.
dict((x, x//16) for x in a if x>32)  # yet another generator exp.
{x//4 for x in a if x>32}            # 2.7+ syntax
{x: x//16 for x in a if x>32}        # 2.7+ syntax

the x is always local to the expression while these:

[x for x in a if x>32]
set([x//4 for x in a if x>32])         # just another list comp.
dict([(x, x//16) for x in a if x>32])  # yet another list comp.

in Python 2.x all leak the x variable to the surrounding scope.

UPDATE for Python 3.8(?): PEP 572 will introduce := assignment operator that deliberately leaks out of comprehensions and generator expressions! It’s motivated by essentially 2 use cases: capturing a “witness” from early-terminating functions like any() and all():

if any((comment := line).startswith('#') for line in lines):
    print("First comment:", comment)
else:
    print("There are no comments")

and updating mutable state:

total = 0
partial_sums = [total := total + v for v in values]

See Appendix B for exact scoping. The variable is assigned in closest surrounding def or lambda, unless that function declares it nonlocal or global.

Question 27

Yes, assignment occurs there, just like it would in a for loop. No new scope is being created.

This is definitely the expected behavior: on each cycle, the value is bound to the name you specify. For instance,

>>> x=0
>>> a=[1,54,4,2,32,234,5234,]
>>> [x for x in a if x>32]
[54, 234, 5234]
>>> x
5234

Once that’s recognized, it seems easy enough to avoid: don’t use existing names for the variables within comprehensions.

Question 28

Interestingly this doesn’t affect dictionary or set comprehensions.

>>> [x for x in range(1, 10)]
[1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> x
9
>>> {x for x in range(1, 5)}
set([1, 2, 3, 4])
>>> x
9
>>> {x:x for x in range(1, 100)}
{1: 1, 2: 2, 3: 3, 4: 4, 5: 5, 6: 6, 7: 7, 8: 8, 9: 9, 10: 10, 11: 11, 12: 12, 13: 13, 14: 14, 15: 15, 16: 16, 17: 17, 18: 18, 19: 19, 20: 20, 21: 21, 22: 22, 23: 23, 24: 24, 25: 25, 26: 26, 27: 27, 28: 28, 29: 29, 30: 30, 31: 31, 32: 32, 33: 33, 34: 34, 35: 35, 36: 36, 37: 37, 38: 38, 39: 39, 40: 40, 41: 41, 42: 42, 43: 43, 44: 44, 45: 45, 46: 46, 47: 47, 48: 48, 49: 49, 50: 50, 51: 51, 52: 52, 53: 53, 54: 54, 55: 55, 56: 56, 57: 57, 58: 58, 59: 59, 60: 60, 61: 61, 62: 62, 63: 63, 64: 64, 65: 65, 66: 66, 67: 67, 68: 68, 69: 69, 70: 70, 71: 71, 72: 72, 73: 73, 74: 74, 75: 75, 76: 76, 77: 77, 78: 78, 79: 79, 80: 80, 81: 81, 82: 82, 83: 83, 84: 84, 85: 85, 86: 86, 87: 87, 88: 88, 89: 89, 90: 90, 91: 91, 92: 92, 93: 93, 94: 94, 95: 95, 96: 96, 97: 97, 98: 98, 99: 99}
>>> x
9

However it has been fixed in 3 as noted above.

Question 29

some workaround, for python 2.6, when this behaviour is not desirable

# python
Python 2.6.6 (r266:84292, Aug  9 2016, 06:11:56)
Type "help", "copyright", "credits" or "license" for more information.
>>> x=0
>>> a=list(x for x in xrange(9))
>>> x
0
>>> a=[x for x in xrange(9)]
>>> x
8

Question 30

In python3 while in list comprehension the variable is not getting change after it’s scope over but when we use simple for-loop the variable is getting reassigned out of scope.

i = 1 print(i) print([i in range(5)]) print(i) Value of i will remain 1 only.

Now just use simply for loop the value of i will be reassigned.

Question 31

How do I rectify the error “unexpected indent” in python?

Question 32

Python uses spacing at the start of the line to determine when code blocks start and end. Errors you can get are:

Unexpected indent. This line of code has more spaces at the start than the one before, but the one before is not the start of a subblock (e.g. if/while/for statement). All lines of code in a block must start with exactly the same string of whitespace. For instance:

>>> def a():
...   print "foo"
...     print "bar"
IndentationError: unexpected indent

This one is especially common when running python interactively: make sure you don’t put any extra spaces before your commands. (Very annoying when copy-and-pasting example code!)

>>>   print "hello"
IndentationError: unexpected indent

Unindent does not match any outer indentation level. This line of code has fewer spaces at the start than the one before, but equally it does not match any other block it could be part of. Python cannot decide where it goes. For instance, in the following, is the final print supposed to be part of the if clause, or not?

>>> if user == "Joey":
...     print "Super secret powers enabled!"
...   print "Revealing super secrets"
IndendationError: unindent does not match any outer indentation level

Expected an indented block. This line of code has the same number of spaces at the start as the one before, but the last line was expected to start a block (e.g. if/while/for statement, function definition).

>>> def foo():
... print "Bar"
IndentationError: expected an indented block

If you want a function that doesn’t do anything, use the “no-op” command pass:

>>> def foo():
...     pass

Mixing tabs and spaces is allowed (at least on my version of Python), but Python assumes tabs are 8 characters long, which may not match your editor. Just say “no” to tabs. Most editors allow them to be automatically replaced by spaces.

The best way to avoid these issues is to always use a consistent number of spaces when you indent a subblock, and ideally use a good IDE that solves the problem for you. This will also make your code more readable.

Question 33

In Python, the spacing is very important, this gives the structure of your code blocks. This error happens when you mess up your code structure, for example like this :

def test_function() :
   if 5 > 3 :
   print "hello"

You may also have a mix of tabs and spaces in your file.

I suggest you use a python syntax aware editor like PyScripter, or Netbeans

Question 34

Run your code with the -tt option to find out if you are using tabs and spaces inconsistently

Question 35

Turn on visible whitespace in whatever editor you are using and turn on replace tabs with spaces.

While you can use tabs with Python mixing tabs and space usually leads to the error you are experiencing. Replacing tabs with 4 spaces is the recommended approach for writing Python code.

Question 36

By using correct indentation. Python is whitespace aware, so you need to follow its indentation guidlines for blocks or you’ll get indentation errors.

Question 37

If you’re writing Python using Sublime and getting indentation errors,

view -> indentation -> convert indentation to spaces

The issue I’m describing is caused by the Sublime text editor. The same issue could be caused by other editors as well. Essentially, the issue has to do with Python wanting to treat indentations in terms of spaces versus various editors coding the indentations in terms of tabs.

Question 38

Make sure you use the option “insert spaces instead of tabs” in your editor. Then you can choose you want a tab width of, for example 4. You can find those options in gedit under edit–>preferences–>editor.

bottom line: USE SPACES not tabs

Question 39

This error can also occur when pasting something into the Python interpreter (terminal/console).

Note that the interpreter interprets an empty line as the end of an expression, so if you paste in something like

def my_function():
    x = 3

    y = 7

the interpreter will interpret the empty line before y = 7 as the end of the expression, i.e. that you’re done defining your function, and the next line – y = 7 will have incorrect indentation because it is a new expression.

Question 40

One issue which doesn’t seem to have been mentioned is that this error can crop up due to a problem with the code that has nothing to do with indentation.

For example, take the following script:

def add_one(x):
    try:
        return x + 1
add_one(5)

This returns an IndentationError: unexpected unindent when the problem is of course a missing except: statement.

My point: check the code above where the unexpected (un)indent is reported!

Question 41

If the indentation looks ok then have a look to see if your editor has a “View Whitespace” option. Enabling this should allow to find where spaces and tabs are mixed.

Question 42

There is a trick that always worked for me:

If you got and unexpected indent and you see that all the code is perfectly indented, try opening it with another editor and you will see what line of code is not indented.

It happened to me when used vim, gedit or editors like that.

Try to use only 1 editor for your code.

Question 43

Simply copy your script and put under “”” your entire code “”” …

specify this line in a variable.. like,

a = """ your python script """
print a.replace('here please press tab button it will insert some space"," here simply press space bar four times")
# here we replacing tab space by four char space as per pep 8 style guide..

now execute this code, in Sublime Editor using ctrl+b, now it will print indented code in console. that’s it

Question 44

All You need to do is remove spaces or tab spaces from the start of following codes

from django.contrib import admin

# Register your models here.
from .models import Myapp
admin.site.register(Myapp)

Question 45

Run the following command to get it solved :

autopep8 -i <filename>.py

This will update your code and solve all indentation Errors :)

Hope this will solve

Question 46

Notepad++ was giving the tab space correct but the indentation problem was finally found in Sublime text editor.

Use Sublime text editor and go line by line

Question 47

Indentation in Python is important and this is just not for code readability, unlike many other programming languages. If there is any white space or tab in your code between consecutive commands, python will give this error as Python is sensitive to this. We are likely to get this error when we do copy and paste of code to any Python. Make sure to identify and remove these spaces using a text editor like Notepad++ or manually remove the whitespace from the line of code where you are getting an error.

Step1 :Gives error 
L = [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]
print(L[2: ])

Step2: L = [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]print(L[2: ])

Step3: No error after space was removed
L = [[1, 2, 3], [4, 5, 6], [7, 8, 9, 10]]
print(L[2: ])
OUTPUT: [[7, 8, 9, 10]]

Thanks!

Question 48

For my django powered site, I am looking for an easy solution to convert dynamic html pages to pdf.

Pages include HTML and charts from Google visualization API (which is javascript based, yet including those graphs is a must).

Question 49

Try the solution from Reportlab.

Download it and install it as usual with python setup.py install

You will also need to install the following modules: xhtml2pdf, html5lib, pypdf with easy_install.

Here is an usage example:

First define this function:

import cStringIO as StringIO
from xhtml2pdf import pisa
from django.template.loader import get_template
from django.template import Context
from django.http import HttpResponse
from cgi import escape


def render_to_pdf(template_src, context_dict):
    template = get_template(template_src)
    context = Context(context_dict)
    html  = template.render(context)
    result = StringIO.StringIO()

    pdf = pisa.pisaDocument(StringIO.StringIO(html.encode("ISO-8859-1")), result)
    if not pdf.err:
        return HttpResponse(result.getvalue(), content_type='application/pdf')
    return HttpResponse('We had some errors<pre>%s</pre>' % escape(html))

Then you can use it like this:

def myview(request):
    #Retrieve data or whatever you need
    return render_to_pdf(
            'mytemplate.html',
            {
                'pagesize':'A4',
                'mylist': results,
            }
        )

The template:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
    <head>
        <title>My Title</title>
        <style type="text/css">
            @page {
                size: {{ pagesize }};
                margin: 1cm;
                @frame footer {
                    -pdf-frame-content: footerContent;
                    bottom: 0cm;
                    margin-left: 9cm;
                    margin-right: 9cm;
                    height: 1cm;
                }
            }
        </style>
    </head>
    <body>
        <div>
            {% for item in mylist %}
                RENDER MY CONTENT
            {% endfor %}
        </div>
        <div id="footerContent">
            {%block page_foot%}
                Page <pdf:pagenumber>
            {%endblock%}
        </div>
    </body>
</html>

Hope it helps.

Question 50

https://github.com/nigma/django-easy-pdf

Template:

{% extends "easy_pdf/base.html" %}

{% block content %}
    <div id="content">
        <h1>Hi there!</h1>
    </div>
{% endblock %}

View:

from easy_pdf.views import PDFTemplateView

class HelloPDFView(PDFTemplateView):
    template_name = "hello.html"

If you want to use django-easy-pdf on Python 3 check the solution suggested here.

Question 51

I just whipped this up for CBV. Not used in production but generates a PDF for me. Probably needs work for the error reporting side of things but does the trick so far.

import StringIO
from cgi import escape
from xhtml2pdf import pisa
from django.http import HttpResponse
from django.template.response import TemplateResponse
from django.views.generic import TemplateView

class PDFTemplateResponse(TemplateResponse):

    def generate_pdf(self, retval):

        html = self.content

        result = StringIO.StringIO()
        rendering = pisa.pisaDocument(StringIO.StringIO(html.encode("ISO-8859-1")), result)

        if rendering.err:
            return HttpResponse('We had some errors<pre>%s</pre>' % escape(html))
        else:
            self.content = result.getvalue()

    def __init__(self, *args, **kwargs):
        super(PDFTemplateResponse, self).__init__(*args, mimetype='application/pdf', **kwargs)
        self.add_post_render_callback(self.generate_pdf)


class PDFTemplateView(TemplateView):
    response_class = PDFTemplateResponse

Used like:

class MyPdfView(PDFTemplateView):
    template_name = 'things/pdf.html'

Question 52

Try wkhtmltopdf with either one of the following wrappers

django-wkhtmltopdf or python-pdfkit

This worked great for me,supports javascript and css or anything for that matter which a webkit browser supports.

For more detailed tutorial please see this blog post

Question 53

After trying to get this to work for too many hours, I finally found this: https://github.com/vierno/django-xhtml2pdf

It’s a fork of https://github.com/chrisglass/django-xhtml2pdf that provides a mixin for a generic class-based view. I used it like this:

    # views.py
    from django_xhtml2pdf.views import PdfMixin
    class GroupPDFGenerate(PdfMixin, DetailView):
        model = PeerGroupSignIn
        template_name = 'groups/pdf.html'

    # templates/groups/pdf.html
    <html>
    <style>
    @page { your xhtml2pdf pisa PDF parameters }
    </style>
    </head>
    <body>
        <div id="header_content"> (this is defined in the style section)
            <h1>{{ peergroupsignin.this_group_title }}</h1>
            ...

Use the model name you defined in your view in all lowercase when populating the template fields. Because its a GCBV, you can just call it as ‘.as_view’ in your urls.py:

    # urls.py (using url namespaces defined in the main urls.py file)
    url(
        regex=r"^(?P<pk>\d+)/generate_pdf/$",
        view=views.GroupPDFGenerate.as_view(),
        name="generate_pdf",
       ),

Question 54

You can use iReport editor to define the layout, and publish the report in jasper reports server. After publish you can invoke the rest api to get the results.

Here is the test of the functionality:

from django.test import TestCase
from x_reports_jasper.models import JasperServerClient

"""
    to try integraction with jasper server through rest
"""
class TestJasperServerClient(TestCase):

    # define required objects for tests
    def setUp(self):

        # load the connection to remote server
        try:

            self.j_url = "http://127.0.0.1:8080/jasperserver"
            self.j_user = "jasperadmin"
            self.j_pass = "jasperadmin"

            self.client = JasperServerClient.create_client(self.j_url,self.j_user,self.j_pass)

        except Exception, e:
            # if errors could not execute test given prerrequisites
            raise

    # test exception when server data is invalid
    def test_login_to_invalid_address_should_raise(self):
        self.assertRaises(Exception,JasperServerClient.create_client, "http://127.0.0.1:9090/jasperserver",self.j_user,self.j_pass)

    # test execute existent report in server
    def test_get_report(self):

        r_resource_path = "/reports/<PathToPublishedReport>"
        r_format = "pdf"
        r_params = {'PARAM_TO_REPORT':"1",}

        #resource_meta = client.load_resource_metadata( rep_resource_path )

        [uuid,out_mime,out_data] = self.client.generate_report(r_resource_path,r_format,r_params)
        self.assertIsNotNone(uuid)

And here is an example of the invocation implementation:

from django.db import models
import requests
import sys
from xml.etree import ElementTree
import logging 

# module logger definition
logger = logging.getLogger(__name__)

# Create your models here.
class JasperServerClient(models.Manager):

    def __handle_exception(self, exception_root, exception_id, exec_info ):
        type, value, traceback = exec_info
        raise JasperServerClientError(exception_root, exception_id), None, traceback

    # 01: REPORT-METADATA 
    #   get resource description to generate the report
    def __handle_report_metadata(self, rep_resourcepath):

        l_path_base_resource = "/rest/resource"
        l_path = self.j_url + l_path_base_resource
        logger.info( "metadata (begin) [path=%s%s]"  %( l_path ,rep_resourcepath) )

        resource_response = None
        try:
            resource_response = requests.get( "%s%s" %( l_path ,rep_resourcepath) , cookies = self.login_response.cookies)

        except Exception, e:
            self.__handle_exception(e, "REPORT_METADATA:CALL_ERROR", sys.exc_info())

        resource_response_dom = None
        try:
            # parse to dom and set parameters
            logger.debug( " - response [data=%s]"  %( resource_response.text) )
            resource_response_dom = ElementTree.fromstring(resource_response.text)

            datum = "" 
            for node in resource_response_dom.getiterator():
                datum = "%s<br />%s - %s" % (datum, node.tag, node.text)
            logger.debug( " - response [xml=%s]"  %( datum ) )

            #
            self.resource_response_payload= resource_response.text
            logger.info( "metadata (end) ")
        except Exception, e:
            logger.error( "metadata (error) [%s]" % (e))
            self.__handle_exception(e, "REPORT_METADATA:PARSE_ERROR", sys.exc_info())


    # 02: REPORT-PARAMS 
    def __add_report_params(self, metadata_text, params ):
        if(type(params) != dict):
            raise TypeError("Invalid parameters to report")
        else:
            logger.info( "add-params (begin) []" )
            #copy parameters
            l_params = {}
            for k,v in params.items():
                l_params[k]=v
            # get the payload metadata
            metadata_dom = ElementTree.fromstring(metadata_text)
            # add attributes to payload metadata
            root = metadata_dom #('report'):

            for k,v in l_params.items():
                param_dom_element = ElementTree.Element('parameter')
                param_dom_element.attrib["name"] = k
                param_dom_element.text = v
                root.append(param_dom_element)

            #
            metadata_modified_text =ElementTree.tostring(metadata_dom, encoding='utf8', method='xml')
            logger.info( "add-params (end) [payload-xml=%s]" %( metadata_modified_text )  )
            return metadata_modified_text



    # 03: REPORT-REQUEST-CALL 
    #   call to generate the report
    def __handle_report_request(self, rep_resourcepath, rep_format, rep_params):

        # add parameters
        self.resource_response_payload = self.__add_report_params(self.resource_response_payload,rep_params)

        # send report request

        l_path_base_genreport = "/rest/report"
        l_path = self.j_url + l_path_base_genreport
        logger.info( "report-request (begin) [path=%s%s]"  %( l_path ,rep_resourcepath) )

        genreport_response = None
        try:
            genreport_response = requests.put( "%s%s?RUN_OUTPUT_FORMAT=%s" %(l_path,rep_resourcepath,rep_format),data=self.resource_response_payload, cookies = self.login_response.cookies )
            logger.info( " - send-operation-result [value=%s]"  %( genreport_response.text) )
        except Exception,e:
            self.__handle_exception(e, "REPORT_REQUEST:CALL_ERROR", sys.exc_info())


        # parse the uuid of the requested report
        genreport_response_dom = None

        try:
            genreport_response_dom = ElementTree.fromstring(genreport_response.text)

            for node in genreport_response_dom.findall("uuid"):
                datum = "%s" % (node.text)

            genreport_uuid = datum      

            for node in genreport_response_dom.findall("file/[@type]"):
                datum = "%s" % (node.text)
            genreport_mime = datum

            logger.info( "report-request (end) [uuid=%s,mime=%s]"  %( genreport_uuid, genreport_mime) )

            return [genreport_uuid,genreport_mime]
        except Exception,e:
            self.__handle_exception(e, "REPORT_REQUEST:PARSE_ERROR", sys.exc_info())

    # 04: REPORT-RETRIEVE RESULTS 
    def __handle_report_reply(self, genreport_uuid ):


        l_path_base_getresult = "/rest/report"
        l_path = self.j_url + l_path_base_getresult 
        logger.info( "report-reply (begin) [uuid=%s,path=%s]"  %( genreport_uuid,l_path) )

        getresult_response = requests.get( "%s%s/%s?file=report" %(self.j_url,l_path_base_getresult,genreport_uuid),data=self.resource_response_payload, cookies = self.login_response.cookies )
        l_result_header_mime =getresult_response.headers['Content-Type']

        logger.info( "report-reply (end) [uuid=%s,mime=%s]"  %( genreport_uuid, l_result_header_mime) )
        return [l_result_header_mime, getresult_response.content]

    # public methods ---------------------------------------    

    # tries the authentication with jasperserver throug rest
    def login(self, j_url, j_user,j_pass):
        self.j_url= j_url

        l_path_base_auth = "/rest/login"
        l_path = self.j_url + l_path_base_auth

        logger.info( "login (begin) [path=%s]"  %( l_path) )

        try:
            self.login_response = requests.post(l_path , params = {
                    'j_username':j_user,
                    'j_password':j_pass
                })                  

            if( requests.codes.ok != self.login_response.status_code ):
                self.login_response.raise_for_status()

            logger.info( "login (end)" )
            return True
            # see http://blog.ianbicking.org/2007/09/12/re-raising-exceptions/

        except Exception, e:
            logger.error("login (error) [e=%s]" % e )
            self.__handle_exception(e, "LOGIN:CALL_ERROR",sys.exc_info())
            #raise

    def generate_report(self, rep_resourcepath,rep_format,rep_params):
        self.__handle_report_metadata(rep_resourcepath)
        [uuid,mime] = self.__handle_report_request(rep_resourcepath, rep_format,rep_params)
        # TODO: how to handle async?
        [out_mime,out_data] = self.__handle_report_reply(uuid)
        return [uuid,out_mime,out_data]

    @staticmethod
    def create_client(j_url, j_user, j_pass):
        client = JasperServerClient()
        login_res = client.login( j_url, j_user, j_pass )
        return client


class JasperServerClientError(Exception):

    def __init__(self,exception_root,reason_id,reason_message=None):
        super(JasperServerClientError, self).__init__(str(reason_message))
        self.code = reason_id 
        self.description = str(exception_root) + " " + str(reason_message)
    def __str__(self):
        return self.code + " " + self.description

Question 55

I get the code to generate the PDF from html template :

    import os

    from weasyprint import HTML

    from django.template import Template, Context
    from django.http import HttpResponse 


    def generate_pdf(self, report_id):

            # Render HTML into memory and get the template firstly
            template_file_loc = os.path.join(os.path.dirname(__file__), os.pardir, 'templates', 'the_template_pdf_generator.html')
            template_contents = read_all_as_str(template_file_loc)
            render_template = Template(template_contents)

            #rendering_map is the dict for params in the template 
            render_definition = Context(rendering_map)
            render_output = render_template.render(render_definition)

            # Using Rendered HTML to generate PDF
            response = HttpResponse(content_type='application/pdf')
            response['Content-Disposition'] = 'attachment; filename=%s-%s-%s.pdf' % \
                                              ('topic-test','topic-test', '2018-05-04')
            # Generate PDF
            pdf_doc = HTML(string=render_output).render()
            pdf_doc.pages[0].height = pdf_doc.pages[0]._page_box.children[0].children[
                0].height  # Make PDF file as single page file 
            pdf_doc.write_pdf(response)
            return response

    def read_all_as_str(self, file_loc, read_method='r'):
        if file_exists(file_loc):
            handler = open(file_loc, read_method)
            contents = handler.read()
            handler.close()
            return contents
        else:
            return 'file not exist'

Question 56

If you have context data along with css and js in your html template. Than you have good option to use pdfjs.

In your code you can use like this.

from django.template.loader import get_template
import pdfkit
from django.conf import settings

context={....}
template = get_template('reports/products.html')
html_string = template.render(context)
pdfkit.from_string(html_string, os.path.join(settings.BASE_DIR, "media", 'products_report-%s.pdf'%(id)))

In your HTML you can link extranal or internal css and js, it will generate best quality of pdf.

Question 57

I am getting the following error while trying to import from sklearn:

>>> from sklearn import svm

Traceback (most recent call last):
  File "<pyshell#17>", line 1, in <module>
   from sklearn import svm
  File "C:\Python27\lib\site-packages\sklearn\__init__.py", line 16, in <module>
   from . import check_build
ImportError: cannot import name check_build

I am using python 2.7, scipy-0.12.0b1 superpack, numpy-1.6.0 superpack, scikit-learn-0.11 I have a windows 7 machine

I have checked several answers for this issue but none of them gives a way out of this error.

Question 58

Worked for me after installing scipy.

Question 59

>>> from sklearn import preprocessing, metrics, cross_validation

Traceback (most recent call last):
  File "<pyshell#6>", line 1, in <module>
    from sklearn import preprocessing, metrics, cross_validation
  File "D:\Python27\lib\site-packages\sklearn\__init__.py", line 31, in <module>
    from . import __check_build
ImportError: cannot import name __check_build
>>> ================================ RESTART ================================
>>> from sklearn import preprocessing, metrics, cross_validation
>>>

So, simply try to restart the shell!

Question 60

My solution for Python 3.6.5 64-bit Windows 10:

pip uninstall sklearn
pip uninstall scikit-learn
pip install sklearn

No need to restart command-line but you can do this if you want. It took me one day to fix this bug. Hope this help.

Question 61

After installing numpy , scipy ,sklearn still has error

Solution:

Setting Up System Path Variable for Python & the PYTHONPATH Environment Variable

System Variables: add C:\Python34 into path User Variables: add new: (name)PYTHONPATH (value)C:\Python34\Lib\site-packages;

Question 62

Usually when I get these kinds of errors, opening the __init__.py file and poking around helps. Go to the directory C:\Python27\lib\site-packages\sklearn and ensure that there’s a sub-directory called __check_build as a first step. On my machine (with a working sklearn installation, Mac OSX, Python 2.7.3) I have __init__.py, setup.py, their associated .pyc files, and a binary _check_build.so.

Poking around the __init__.py in that directory, the next step I’d take is to go to sklearn/__init__.py and comment out the import statement—the check_build stuff just checks that things were compiled correctly, it doesn’t appear to do anything but call a precompiled binary. This is, of course, at your own risk, and (to be sure) a work around. If your build failed you’ll likely soon run into other, bigger problems.

Question 63

I had the same issue on Windows. Solved it by installing Numpy+MKL from http://www.lfd.uci.edu/~gohlke/pythonlibs/#numpy (there it’s recommended to install numpy+mkl before other packages that depend on it) as suggested by this answer.

Question 64

I had problems importing SKLEARN after installing a new 64bit version of Python 3.4 from python.org.

Turns out that it was the SCIPY module that was broken, and alos failed when I tried to “import scipy”.

Solution was to uninstall scipy and reinstall it with pip3:

C:\> pip uninstall scipy

[lots of reporting messages deleted]

Proceed (y/n)? y
  Successfully uninstalled scipy-1.0.0

C:\Users\>pip3 install scipy

Collecting scipy
  Downloading scipy-1.0.0-cp36-none-win_amd64.whl (30.8MB)
    100% |████████████████████████████████| 30.8MB 33kB/s
Requirement already satisfied: numpy>=1.8.2 in c:\users\johnmccurdy\appdata\loca
l\programs\python\python36\lib\site-packages (from scipy)
Installing collected packages: scipy
Successfully installed scipy-1.0.0

C:\Users>python
Python 3.6.4 (v3.6.4:d48eceb, Dec 19 2017, 06:54:40) [MSC v.1900 64 bit (AMD64)]
 on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> import scipy
>>>
>>> import sklearn
>>>

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