Python 实用宝典

Question 1

I’ve got the following simple script that plots a graph:

import matplotlib.pyplot as plt
import numpy as np

T = np.array([6, 7, 8, 9, 10, 11, 12])
power = np.array([1.53E+03, 5.92E+02, 2.04E+02, 7.24E+01, 2.72E+01, 1.10E+01, 4.70E+00])

plt.plot(T,power)
plt.show()

As it is now, the line goes straight from point to point which looks ok, but could be better in my opinion. What I want is to smooth the line between the points. In Gnuplot I would have plotted with smooth cplines.

Is there an easy way to do this in PyPlot? I’ve found some tutorials, but they all seem rather complex.

Question 2

You could use scipy.interpolate.spline to smooth out your data yourself:

from scipy.interpolate import spline

# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300)  

power_smooth = spline(T, power, xnew)

plt.plot(xnew,power_smooth)
plt.show()

spline is deprecated in scipy 0.19.0, use BSpline class instead.

Switching from spline to BSpline isn’t a straightforward copy/paste and requires a little tweaking:

from scipy.interpolate import make_interp_spline, BSpline

# 300 represents number of points to make between T.min and T.max
xnew = np.linspace(T.min(), T.max(), 300) 

spl = make_interp_spline(T, power, k=3)  # type: BSpline
power_smooth = spl(xnew)

plt.plot(xnew, power_smooth)
plt.show()

Before:

After:

Question 3

For this example spline works well, but if the function is not smooth inherently and you want to have smoothed version you can also try:

from scipy.ndimage.filters import gaussian_filter1d

ysmoothed = gaussian_filter1d(y, sigma=2)
plt.plot(x, ysmoothed)
plt.show()

if you increase sigma you can get a more smoothed function.

Proceed with caution with this one. It modifies the original values and may not be what you want.

Question 4

I presume you mean curve-fitting and not anti-aliasing from the context of your question. PyPlot doesn’t have any built-in support for this, but you can easily implement some basic curve-fitting yourself, like the code seen here, or if you’re using GuiQwt it has a curve fitting module. (You could probably also steal the code from SciPy to do this as well).

Question 5

See the scipy.interpolate documentation for some examples.

The following example demonstrates its use, for linear and cubic spline interpolation:
>>> from scipy.interpolate import interp1d

>>> x = np.linspace(0, 10, num=11, endpoint=True)
>>> y = np.cos(-x**2/9.0)
>>> f = interp1d(x, y)
>>> f2 = interp1d(x, y, kind='cubic')

>>> xnew = np.linspace(0, 10, num=41, endpoint=True)
>>> import matplotlib.pyplot as plt
>>> plt.plot(x, y, 'o', xnew, f(xnew), '-', xnew, f2(xnew), '--')
>>> plt.legend(['data', 'linear', 'cubic'], loc='best')
>>> plt.show()

Question 6

Based on this question about heatmaps in matplotlib, I wanted to move the x-axis titles to the top of the plot.

import matplotlib.pyplot as plt
import numpy as np
column_labels = list('ABCD')
row_labels = list('WXYZ')
data = np.random.rand(4,4)
fig, ax = plt.subplots()
heatmap = ax.pcolor(data, cmap=plt.cm.Blues)

# put the major ticks at the middle of each cell
ax.set_xticks(np.arange(data.shape[0])+0.5, minor=False)
ax.set_yticks(np.arange(data.shape[1])+0.5, minor=False)

# want a more natural, table-like display
ax.invert_yaxis()
ax.xaxis.set_label_position('top') # <-- This doesn't work!

ax.set_xticklabels(row_labels, minor=False)
ax.set_yticklabels(column_labels, minor=False)
plt.show()

However, calling matplotlib’s set_label_position (as notated above) doesn’t seem to have the desired effect. Here’s my output:

What am I doing wrong?

Question 7

Use

ax.xaxis.tick_top()

to place the tick marks at the top of the image. The command

ax.set_xlabel('X LABEL')    
ax.xaxis.set_label_position('top')

affects the label, not the tick marks.

import matplotlib.pyplot as plt
import numpy as np
column_labels = list('ABCD')
row_labels = list('WXYZ')
data = np.random.rand(4, 4)
fig, ax = plt.subplots()
heatmap = ax.pcolor(data, cmap=plt.cm.Blues)

# put the major ticks at the middle of each cell
ax.set_xticks(np.arange(data.shape[1]) + 0.5, minor=False)
ax.set_yticks(np.arange(data.shape[0]) + 0.5, minor=False)

# want a more natural, table-like display
ax.invert_yaxis()
ax.xaxis.tick_top()

ax.set_xticklabels(column_labels, minor=False)
ax.set_yticklabels(row_labels, minor=False)
plt.show()

Question 8

You want set_ticks_position rather than set_label_position:

ax.xaxis.set_ticks_position('top') # the rest is the same

This gives me:

Question 9

tick_params is very useful for setting tick properties. Labels can be moved to the top with:

    ax.tick_params(labelbottom=False,labeltop=True)

Question 10

You’ve got to do some extra massaging if you want the ticks (not labels) to show up on the top and bottom (not just the top). The only way I could do this is with a minor change to unutbu’s code:

import matplotlib.pyplot as plt
import numpy as np
column_labels = list('ABCD')
row_labels = list('WXYZ')
data = np.random.rand(4, 4)
fig, ax = plt.subplots()
heatmap = ax.pcolor(data, cmap=plt.cm.Blues)

# put the major ticks at the middle of each cell
ax.set_xticks(np.arange(data.shape[1]) + 0.5, minor=False)
ax.set_yticks(np.arange(data.shape[0]) + 0.5, minor=False)

# want a more natural, table-like display
ax.invert_yaxis()
ax.xaxis.tick_top()
ax.xaxis.set_ticks_position('both') # THIS IS THE ONLY CHANGE

ax.set_xticklabels(column_labels, minor=False)
ax.set_yticklabels(row_labels, minor=False)
plt.show()

Output:

Question 11

I have a python script that requires some command line inputs and I am using argparse for parsing them. I found the documentation a bit confusing and couldn’t find a way to check for a format in the input parameters. What I mean by checking format is explained with this example script:

parser.add_argument('-s', "--startdate", help="The Start Date - format YYYY-MM-DD ", required=True)
parser.add_argument('-e', "--enddate", help="The End Date format YYYY-MM-DD (Inclusive)", required=True)
parser.add_argument('-a', "--accountid", type=int, help='Account ID for the account for which data is required (Default: 570)')
parser.add_argument('-o', "--outputpath", help='Directory where output needs to be stored (Default: ' + os.path.dirname(os.path.abspath(__file__)))

I need to check for option -s and -e that the input by the user is in the format YYYY-MM-DD. Is there an option in argparse that I do not know of which accomplishes this.

Question 12

Per the documentation:

The type keyword argument of add_argument() allows any necessary type-checking and type conversions to be performed … type= can take any callable that takes a single string argument and returns the converted value

You could do something like:

def valid_date(s):
    try:
        return datetime.strptime(s, "%Y-%m-%d")
    except ValueError:
        msg = "Not a valid date: '{0}'.".format(s)
        raise argparse.ArgumentTypeError(msg)

Then use that as type:

parser.add_argument("-s", 
                    "--startdate", 
                    help="The Start Date - format YYYY-MM-DD", 
                    required=True, 
                    type=valid_date)

Question 13

Just to add on to the answer above, you can use a lambda function if you want to keep it to a one-liner. For example:

parser.add_argument('--date', type=lambda d: datetime.strptime(d, '%Y%m%d'))

Old thread but the question was still relevant for me at least!

Question 14

For others who hit this via search engines: in Python 3.7, you can use the standard .fromisoformat class method instead of reinventing the wheel for ISO-8601 compliant dates, e.g.:

parser.add_argument('-s', "--startdate",
    help="The Start Date - format YYYY-MM-DD",
    required=True,
    type=datetime.date.fromisoformat)
parser.add_argument('-e', "--enddate",
    help="The End Date format YYYY-MM-DD (Inclusive)",
    required=True,
    type=datetime.date.fromisoformat)

Question 15

I have commented out csrf processor and middleware lines in settings.py:

122 
123 TEMPLATE_CONTEXT_PROCESSORS = (
124     'django.contrib.auth.context_processors.auth',
125 #    'django.core.context_processors.csrf',
126     'django.core.context_processors.request',
127     'django.core.context_processors.static',
128     'cyathea.processors.static',
129 )
130 
131 MIDDLEWARE_CLASSES = (
132     'django.middleware.common.CommonMiddleware',
133     'django.contrib.sessions.middleware.SessionMiddleware',
134 #    'django.middleware.csrf.CsrfViewMiddleware',
135     'django.contrib.auth.middleware.AuthenticationMiddleware',
136     'django.contrib.messages.middleware.MessageMiddleware',
137     'django.middleware.locale.LocaleMiddleware',
138     # Uncomment the next line for simple clickjacking protection:
139     # 'django.middleware.clickjacking.XFrameOptionsMiddleware',
140 )

But when I use Ajax to send a request, Django still respond ‘csrf token is incorrect or missing’, and after adding X-CSRFToken to headers, the request would succeed.

What is going on here ?

Question 16

If you just need some views not to use CSRF, you can use @csrf_exempt:

from django.views.decorators.csrf import csrf_exempt

@csrf_exempt
def my_view(request):
    return HttpResponse('Hello world')

You can find more examples and other scenarios in the Django documentation:

https://docs.djangoproject.com/en/dev/ref/csrf/#edge-cases

Question 17

To disable CSRF for class based views the following worked for me.
Using django 1.10 and python 3.5.2

from django.views.decorators.csrf import csrf_exempt
from django.utils.decorators import method_decorator

@method_decorator(csrf_exempt, name='dispatch')
class TestView(View):
    def post(self, request, *args, **kwargs):
        return HttpResponse('Hello world')

Question 18

In setting.py in MIDDLEWARE you can simply remove/comment this line:

'django.middleware.csrf.CsrfViewMiddleware',

Question 19

For Django 2:

from django.utils.deprecation import MiddlewareMixin


class DisableCSRF(MiddlewareMixin):
    def process_request(self, request):
        setattr(request, '_dont_enforce_csrf_checks', True)

That middleware must be added to settings.MIDDLEWARE when appropriate (in your test settings for example).

Note: the setting isn’t not called MIDDLEWARE_CLASSES anymore.

Question 20

The answer might be inappropriate, but I hope it helps you

class DisableCSRFOnDebug(object):
    def process_request(self, request):
        if settings.DEBUG:
            setattr(request, '_dont_enforce_csrf_checks', True)

Having middleware like this helps to debug requests and to check csrf in production servers.

Question 21

The problem here is that SessionAuthentication performs its own CSRF validation. That is why you get the CSRF missing error even when the CSRF Middleware is commented. You could add @csrf_exempt to every view, but if you want to disable CSRF and have session authentication for the whole app, you can add an extra middleware like this –

class DisableCSRFMiddleware(object):

def __init__(self, get_response):
    self.get_response = get_response

def __call__(self, request):
    setattr(request, '_dont_enforce_csrf_checks', True)
    response = self.get_response(request)
    return response

I created this class in myapp/middle.py Then import this middleware in Middleware in settings.py

MIDDLEWARE = [
    'django.middleware.common.CommonMiddleware',
    'django.middleware.security.SecurityMiddleware',
    'django.contrib.sessions.middleware.SessionMiddleware',
    'django.middleware.common.CommonMiddleware',
    #'django.middleware.csrf.CsrfViewMiddleware',
    'myapp.middle.DisableCSRFMiddleware',
    'django.contrib.auth.middleware.AuthenticationMiddleware',
    'django.contrib.messages.middleware.MessageMiddleware',
    'django.middleware.clickjacking.XFrameOptionsMiddleware',

]

That works with DRF on django 1.11

Question 22

If you want disable it in Global, you can write a custom middleware, like this

from django.utils.deprecation import MiddlewareMixin

class DisableCsrfCheck(MiddlewareMixin):

    def process_request(self, req):
        attr = '_dont_enforce_csrf_checks'
        if not getattr(req, attr, False):
            setattr(req, attr, True)

then add this class youappname.middlewarefilename.DisableCsrfCheck to MIDDLEWARE_CLASSES lists, before django.middleware.csrf.CsrfViewMiddleware

Question 23

CSRF can be enforced at the view level, which can’t be disabled globally.

In some cases this is a pain, but um, “it’s for security”. Gotta retain those AAA ratings.

https://docs.djangoproject.com/en/dev/ref/csrf/#contrib-and-reusable-apps

Question 24

@WoooHaaaa some third party packages use ‘django.middleware.csrf.CsrfViewMiddleware’ middleware. for example i use django-rest-oauth and i have problem like you even after disabling those things. maybe these packages responded to your request like my case, because you use authentication decorator and something like this.

Question 25

This is a copy of someone else’s question on another forum that was never answered, so I thought I’d re-ask it here, as I have the same issue. (See http://geekple.com/blogs/feeds/Xgzu7/posts/351703064084736)

I have Spark installed properly on my machine and am able to run python programs with the pyspark modules without error when using ./bin/pyspark as my python interpreter.

However, when I attempt to run the regular Python shell, when I try to import pyspark modules I get this error:

from pyspark import SparkContext

and it says

"No module named pyspark".

How can I fix this? Is there an environment variable I need to set to point Python to the pyspark headers/libraries/etc.? If my spark installation is /spark/, which pyspark paths do I need to include? Or can pyspark programs only be run from the pyspark interpreter?

Question 26

Here is a simple method (If you don’t bother about how it works!!!)

Use findspark

Go to your python shell

pip install findspark

import findspark
findspark.init()

import the necessary modules

from pyspark import SparkContext
from pyspark import SparkConf

Done!!!

Question 27

If it prints such error:

ImportError: No module named py4j.java_gateway

Please add $SPARK_HOME/python/build to PYTHONPATH:

export SPARK_HOME=/Users/pzhang/apps/spark-1.1.0-bin-hadoop2.4
export PYTHONPATH=$SPARK_HOME/python:$SPARK_HOME/python/build:$PYTHONPATH

Question 28

Turns out that the pyspark bin is LOADING python and automatically loading the correct library paths. Check out $SPARK_HOME/bin/pyspark :

# Add the PySpark classes to the Python path:
export PYTHONPATH=$SPARK_HOME/python/:$PYTHONPATH

I added this line to my .bashrc file and the modules are now correctly found!

Question 29

dont run your py file as: python filename.py instead use: spark-submit filename.py

Question 30

By exporting the SPARK path and the Py4j path, it started to work:

export SPARK_HOME=/usr/local/Cellar/apache-spark/1.5.1
export PYTHONPATH=$SPARK_HOME/libexec/python:$SPARK_HOME/libexec/python/build:$PYTHONPATH
PYTHONPATH=$SPARK_HOME/python/lib/py4j-0.8.2.1-src.zip:$PYTHONPATH 
export PYTHONPATH=$SPARK_HOME/python:$SPARK_HOME/python/build:$PYTHONPATH

So, if you don’t want to type these everytime you want to fire up the Python shell, you might want to add it to your .bashrc file

Question 31

On Mac, I use Homebrew to install Spark (formula “apache-spark”). Then, I set the PYTHONPATH this way so the Python import works:

export SPARK_HOME=/usr/local/Cellar/apache-spark/1.2.0
export PYTHONPATH=$SPARK_HOME/libexec/python:$SPARK_HOME/libexec/python/build:$PYTHONPATH

Replace the “1.2.0” with the actual apache-spark version on your mac.

Question 32

For a Spark execution in pyspark two components are required to work together:

pyspark python package
Spark instance in a JVM

When launching things with spark-submit or pyspark, these scripts will take care of both, i.e. they set up your PYTHONPATH, PATH, etc, so that your script can find pyspark, and they also start the spark instance, configuring according to your params, e.g. –master X

Alternatively, it is possible to bypass these scripts and run your spark application directly in the python interpreter likepython myscript.py. This is especially interesting when spark scripts start to become more complex and eventually receive their own args.

Ensure the pyspark package can be found by the Python interpreter. As already discussed either add the spark/python dir to PYTHONPATH or directly install pyspark using pip install.
Set the parameters of spark instance from your script (those that used to be passed to pyspark).
- For spark configurations as you’d normally set with –conf they are defined with a config object (or string configs) in SparkSession.builder.config
- For main options (like –master, or –driver-mem) for the moment you can set them by writing to the PYSPARK_SUBMIT_ARGS environment variable. To make things cleaner and safer you can set it from within Python itself, and spark will read it when starting.
Start the instance, which just requires you to call getOrCreate() from the builder object.

Your script can therefore have something like this:

from pyspark.sql import SparkSession

if __name__ == "__main__":
    if spark_main_opts:
        # Set main options, e.g. "--master local[4]"
        os.environ['PYSPARK_SUBMIT_ARGS'] = spark_main_opts + " pyspark-shell"

    # Set spark config
    spark = (SparkSession.builder
             .config("spark.checkpoint.compress", True)
             .config("spark.jars.packages", "graphframes:graphframes:0.5.0-spark2.1-s_2.11")
             .getOrCreate())

Question 33

To get rid of ImportError: No module named py4j.java_gateway, you need to add following lines:

import os
import sys


os.environ['SPARK_HOME'] = "D:\python\spark-1.4.1-bin-hadoop2.4"


sys.path.append("D:\python\spark-1.4.1-bin-hadoop2.4\python")
sys.path.append("D:\python\spark-1.4.1-bin-hadoop2.4\python\lib\py4j-0.8.2.1-src.zip")

try:
    from pyspark import SparkContext
    from pyspark import SparkConf

    print ("success")

except ImportError as e:
    print ("error importing spark modules", e)
    sys.exit(1)

Question 34

On Windows 10 the following worked for me. I added the following environment variables using Settings > Edit environment variables for your account:

SPARK_HOME=C:\Programming\spark-2.0.1-bin-hadoop2.7
PYTHONPATH=%SPARK_HOME%\python;%PYTHONPATH%

(change “C:\Programming\…” to the folder in which you have installed spark)

Question 35

For Linux users, the following is the correct (and non-hard-coded) way of including the pyspark libaray in PYTHONPATH. Both PATH parts are necessary:

The path to the pyspark Python module itself, and
The path to the zipped library that that pyspark module relies on when imported

Notice below that the zipped library version is dynamically determined, so we do not hard-code it.

export PYTHONPATH=${SPARK_HOME}/python/:$(echo ${SPARK_HOME}/python/lib/py4j-*-src.zip):${PYTHONPATH}

Question 36

I am running a spark cluster, on CentOS VM, which is installed from cloudera yum packages.

Had to set the following variables to run pyspark.

export SPARK_HOME=/usr/lib/spark;
export PYTHONPATH=$SPARK_HOME/python:$SPARK_HOME/python/lib/py4j-0.9-src.zip:$PYTHONPATH

Question 37

export PYSPARK_PYTHON=/home/user/anaconda3/bin/python
export PYSPARK_DRIVER_PYTHON=jupyter
export PYSPARK_DRIVER_PYTHON_OPTS='notebook'

This is what I did for using my Anaconda distribution with Spark. This is Spark version independent. You can change the first line to your users’ python bin. Also, as of Spark 2.2.0 PySpark is available as a Stand-alone package on PyPi but I am yet to test it out.

Question 38

You can get the pyspark path in python using pip (if you have installed pyspark using PIP) as below

pip show pyspark

Question 39

I had the same problem.

Also make sure you are using right python version and you are installing it with right pip version. in my case: I had both python 2.7 and 3.x. I have installed pyspark with

pip2.7 install pyspark

and it worked.

Question 40

I got this error because the python script I was trying to submit was called pyspark.py (facepalm). The fix was to set my PYTHONPATH as recommended above, then rename the script to pyspark_test.py and clean up the pyspark.pyc that was created based on my scripts original name and that cleared this error up.

Question 41

In the case of DSE (DataStax Cassandra & Spark) The following location needs to be added to PYTHONPATH

export PYTHONPATH=/usr/share/dse/resources/spark/python:$PYTHONPATH

Then use the dse pyspark to get the modules in path.

dse pyspark

Question 42

I had this same problem and would add one thing to the proposed solutions above. When using Homebrew on Mac OS X to install Spark you will need to correct the py4j path address to include libexec in the path (remembering to change py4j version to the one you have);

PYTHONPATH=$SPARK_HOME/libexec/python/lib/py4j-0.9-src.zip:$PYTHONPATH

Question 43

In my case it was getting install at a different python dist_package (python 3.5) whereas I was using python 3.6, so the below helped:

python -m pip install pyspark

Question 44

You can also create a Docker container with Alpine as the OS and the install Python and Pyspark as packages. That will have it all containerised.

Question 45

Given an Exception object (of unknown origin) is there way to obtain its traceback? I have code like this:

def stuff():
   try:
       .....
       return useful
   except Exception as e:
       return e

result = stuff()
if isinstance(result, Exception):
    result.traceback <-- How?

How can I extract the traceback from the Exception object once I have it?

Question 46

The answer to this question depends on the version of Python you’re using.

In Python 3

It’s simple: exceptions come equipped with a __traceback__ attribute that contains the traceback. This attribute is also writable, and can be conveniently set using the with_traceback method of exceptions:

raise Exception("foo occurred").with_traceback(tracebackobj)

These features are minimally described as part of the raise documentation.

All credit for this part of the answer should go to Vyctor, who first posted this information. I’m including it here only because this answer is stuck at the top, and Python 3 is becoming more common.

In Python 2

It’s annoyingly complex. The trouble with tracebacks is that they have references to stack frames, and stack frames have references to the tracebacks that have references to stack frames that have references to… you get the idea. This causes problems for the garbage collector. (Thanks to ecatmur for first pointing this out.)

The nice way of solving this would be to surgically break the cycle after leaving the except clause, which is what Python 3 does. The Python 2 solution is much uglier: you are provided with an ad-hoc function,sys.exc_info(), which only works inside the except clause. It returns a tuple containing the exception, the exception type, and the traceback for whatever exception is currently being handled.

So if you are inside the except clause, you can use the output of sys.exc_info() along with the traceback module to do various useful things:

>>> import sys, traceback
>>> def raise_exception():
...     try:
...         raise Exception
...     except Exception:
...         ex_type, ex, tb = sys.exc_info()
...         traceback.print_tb(tb)
...     finally:
...         del tb
... 
>>> raise_exception()
  File "<stdin>", line 3, in raise_exception

But as your edit indicates, you’re trying to get the traceback that would have been printed if your exception had not been handled, after it has already been handled. That’s a much harder question. Unfortunately, sys.exc_info returns (None, None, None) when no exception is being handled. Other related sys attributes don’t help either. sys.exc_traceback is deprecated and undefined when no exception is being handled; sys.last_traceback seems perfect, but it appears only to be defined during interactive sessions.

If you can control how the exception is raised, you might be able to use inspect and a custom exception to store some of the information. But I’m not entirely sure how that would work.

To tell the truth, catching and returning an exception is kind of an unusual thing to do. This might be a sign that you need to refactor anyway.

Question 47

Since Python 3.0^{[PEP 3109]} the built in class Exception has a __traceback__ attribute which contains a traceback object (with Python 3.2.3):

>>> try:
...     raise Exception()
... except Exception as e:
...     tb = e.__traceback__
...
>>> tb
<traceback object at 0x00000000022A9208>

The problem is that after Googling __traceback__ for a while I found only few articles but none of them describes whether or why you should (not) use __traceback__.

However, the Python 3 documentation for raise says that:

A traceback object is normally created automatically when an exception is raised and attached to it as the __traceback__ attribute, which is writable.

So I assume it’s meant to be used.

Question 48

A way to get traceback as a string from an exception object in Python 3:

import traceback

# `e` is an exception object that you get from somewhere
traceback_str = ''.join(traceback.format_tb(e.__traceback__))

traceback.format_tb(...) returns a list of strings. ''.join(...) joins them together. For more reference, please visit: https://docs.python.org/3/library/traceback.html#traceback.format_tb

Question 49

As an aside, if you want to actually get the full traceback as you would see it printed to your terminal, you want this:

>>> try:
...     print(1/0)
... except Exception as e:
...     exc = e
...
>>> exc
ZeroDivisionError('division by zero')
>>> tb_str = traceback.format_exception(etype=type(exc), value=exc, tb=exc.__traceback__)
>>> tb_str
['Traceback (most recent call last):\n', '  File "<stdin>", line 2, in <module>\n', 'ZeroDivisionError: division by zero\n']
>>> print("".join(tb_str))
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
ZeroDivisionError: division by zero

If you use format_tb as above answers suggest you’ll get less information:

>>> tb_str = "".join(traceback.format_tb(exc.__traceback__))
>>> print("".join(tb_str))
  File "<stdin>", line 2, in <module>

Question 50

There’s a very good reason the traceback is not stored in the exception; because the traceback holds references to its stack’s locals, this would result in a circular reference and (temporary) memory leak until the circular GC kicks in. (This is why you should never store the traceback in a local variable.)

About the only thing I can think of would be for you to monkeypatch stuff‘s globals so that when it thinks it’s catching Exception it’s actually catching a specialised type and the exception propagates to you as the caller:

module_containing_stuff.Exception = type("BogusException", (Exception,), {})
try:
    stuff()
except Exception:
    import sys
    print sys.exc_info()

Question 51

what I’m trying to do is this:

get the 30 Authors with highest score ( Author.objects.order_by('-score')[:30] )
order the authors by last_name

Any suggestions?

Question 52

What about

import operator

auths = Author.objects.order_by('-score')[:30]
ordered = sorted(auths, key=operator.attrgetter('last_name'))

In Django 1.4 and newer you can order by providing multiple fields.
Reference: https://docs.djangoproject.com/en/dev/ref/models/querysets/#order-by

order_by(*fields)

By default, results returned by a QuerySet are ordered by the ordering tuple given by the ordering option in the model’s Meta. You can override this on a per-QuerySet basis by using the order_by method.

Example:

ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]

The result above will be ordered by score descending, then by last_name ascending. The negative sign in front of "-score" indicates descending order. Ascending order is implied.

Question 53

I just wanted to illustrate that the built-in solutions (SQL-only) are not always the best ones. At first I thought that because Django’s QuerySet.objects.order_by method accepts multiple arguments, you could easily chain them:

ordered_authors = Author.objects.order_by('-score', 'last_name')[:30]

But, it does not work as you would expect. Case in point, first is a list of presidents sorted by score (selecting top 5 for easier reading):

>>> auths = Author.objects.order_by('-score')[:5]
>>> for x in auths: print x
... 
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)

Using Alex Martelli’s solution which accurately provides the top 5 people sorted by last_name:

>>> for x in sorted(auths, key=operator.attrgetter('last_name')): print x
... 
Benjamin Harrison (467)
James Monroe (487)
Gerald Rudolph (464)
Ulysses Simpson (474)
Harry Truman (471)

And now the combined order_by call:

>>> myauths = Author.objects.order_by('-score', 'last_name')[:5]
>>> for x in myauths: print x
... 
James Monroe (487)
Ulysses Simpson (474)
Harry Truman (471)
Benjamin Harrison (467)
Gerald Rudolph (464)

As you can see it is the same result as the first one, meaning it doesn’t work as you would expect.

Question 54

Here’s a way that allows for ties for the cut-off score.

author_count = Author.objects.count()
cut_off_score = Author.objects.order_by('-score').values_list('score')[min(30, author_count)]
top_authors = Author.objects.filter(score__gte=cut_off_score).order_by('last_name')

You may get more than 30 authors in top_authors this way and the min(30,author_count) is there incase you have fewer than 30 authors.

Question 55

I construct a string s in Python 2.6.5 which will have a varying number of %s tokens, which match the number of entries in list x. I need to write out a formatted string. The following doesn’t work, but indicates what I’m trying to do. In this example, there are three %s tokens and the list has three entries.

s = '%s BLAH %s FOO %s BAR'
x = ['1', '2', '3']
print s % (x)

I’d like the output string to be:

1 BLAH 2 FOO 3 BAR

Question 56

print s % tuple(x)

instead of

print s % (x)

Question 57

You should take a look to the format method of python. You could then define your formatting string like this :

>>> s = '{0} BLAH BLAH {1} BLAH {2} BLAH BLIH BLEH'
>>> x = ['1', '2', '3']
>>> print s.format(*x)
'1 BLAH BLAH 2 BLAH 3 BLAH BLIH BLEH'

Question 58

Following this resource page, if the length of x is varying, we can use:

', '.join(['%.2f']*len(x))

to create a place holder for each element from the list x. Here is the example:

x = [1/3.0, 1/6.0, 0.678]
s = ("elements in the list are ["+', '.join(['%.2f']*len(x))+"]") % tuple(x)
print s
>>> elements in the list are [0.33, 0.17, 0.68]

Question 59

Here is a one liner. A little improvised answer using format with print() to iterate a list.

How about this (python 3.x):

sample_list = ['cat', 'dog', 'bunny', 'pig']
print("Your list of animals are: {}, {}, {} and {}".format(*sample_list))

Read the docs here on using format().

Question 60

Since I just learned about this cool thing(indexing into lists from within a format string) I’m adding to this old question.

s = '{x[0]} BLAH {x[1]} FOO {x[2]} BAR'
x = ['1', '2', '3']
print (s.format (x=x))

Output:

1 BLAH 2 FOO 3 BAR

However, I still haven’t figured out how to do slicing(inside of the format string '"{x[2:4]}".format...,) and would love to figure it out if anyone has an idea, however I suspect that you simply cannot do that.

Question 61

This was a fun question! Another way to handle this for variable length lists is to build a function that takes full advantage of the .format method and list unpacking. In the following example I don’t use any fancy formatting, but that can easily be changed to suit your needs.

list_1 = [1,2,3,4,5,6]
list_2 = [1,2,3,4,5,6,7,8]

# Create a function that can apply formatting to lists of any length:
def ListToFormattedString(alist):
    # Create a format spec for each item in the input `alist`.
    # E.g., each item will be right-adjusted, field width=3.
    format_list = ['{:>3}' for item in alist] 

    # Now join the format specs into a single string:
    # E.g., '{:>3}, {:>3}, {:>3}' if the input list has 3 items.
    s = ','.join(format_list)

    # Now unpack the input list `alist` into the format string. Done!
    return s.format(*alist)

# Example output:
>>>ListToFormattedString(list_1)
'  1,  2,  3,  4,  5,  6'
>>>ListToFormattedString(list_2)
'  1,  2,  3,  4,  5,  6,  7,  8'

Question 62

The same as @neobot’s answer but a little more modern and succinct.

>>> l = range(5)
>>> " & ".join(["{}"]*len(l)).format(*l)
'0 & 1 & 2 & 3 & 4'

Question 63

x = ['1', '2', '3']
s = f"{x[0]} BLAH {x[1]} FOO {x[2]} BAR"
print(s)

The output is

1 BLAH 2 FOO 3 BAR

Question 64

I have a string of this form

s='arbit'
string='%s hello world %s hello world %s' %(s,s,s)

All the %s in string have the same value (i.e. s). Is there a better way of writing this? (Rather than listing out s three times)

问题：用PyPlot绘制平滑线

回答 0

回答 1

回答 2

回答 3

问题：在matplotlib中将x轴移动到绘图的顶部

回答 0

回答 1

回答 2

回答 3

问题：指定输入参数argparse python的格式

回答 0

回答 1

回答 2

问题：如何禁用Django的CSRF验证？

回答 0

回答 1

回答 2

回答 3

回答 4

回答 5

回答 6

回答 7

回答 8

问题：在python shell中导入pyspark

回答 0

回答 1

回答 2

回答 3

回答 4

回答 5

回答 6

回答 7

回答 8

回答 9

回答 10

回答 11

回答 12

回答 13

回答 14

回答 15

回答 16

回答 17

回答 18

问题：从异常对象中提取回溯信息

回答 0

在Python 3中

在Python 2中

In Python 3

In Python 2

回答 1

回答 2

回答 3

回答 4

问题：排序查询集的好方法？-Django

回答 0

回答 1

回答 2

问题：将Python字符串格式化与列表一起使用

回答 0

回答 1

回答 2

回答 3

回答 4

回答 5

回答 6

回答 7

问题：格式化字符串时多次插入相同的值

回答 0

回答 1

回答 2

回答 3

回答 4

回答 5

弦线

基准测试

Fstrings

Benchmarks

问题：Python可以打印函数定义吗？

回答 0