Max retries exceeded with URL in requests

| |

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I"m trying to get the content of App Store > Business:

import requests
from lxml import html

page = requests.get("")
tree = html.fromstring(page.text)

flist = []
plist = []
for i in range(0, 100):
    app = tree.xpath("//div[@class="column first"]/ul/li/a/@href")
    ap = app[0]
    page1 = requests.get(ap)

When I try the range with (0,2) it works, but when I put the range in 100s it shows this error:

Traceback (most recent call last):
  File "/home/preetham/Desktop/", line 17, in <module>
    page1 = requests.get(ap)
  File "/usr/local/lib/python2.7/dist-packages/requests/", line 55, in get
    return request("get", url, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/requests/", line 44, in request
    return session.request(method=method, url=url, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/requests/", line 383, in request
    resp = self.send(prep, **send_kwargs)
  File "/usr/local/lib/python2.7/dist-packages/requests/", line 486, in send
    r = adapter.send(request, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/requests/", line 378, in send
    raise ConnectionError(e)
requests.exceptions.ConnectionError: HTTPSConnectionPool(host="", port=443): Max retries exceeded with url: /in/app/adobe-reader/id469337564?mt=8 (Caused by <class "socket.gaierror">: [Errno -2] Name or service not known)

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Max retries exceeded with URL in requests sin: Questions


How do I merge two dictionaries in a single expression (taking union of dictionaries)?

5 answers

Carl Meyer By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)


Answer #1

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
  • In Python 3.5 or greater:

    z = {**x, **y}
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z

    and now:

    z = merge_two_dicts(x, y)


Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    result = {}
    for dictionary in dict_args:
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.


Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z


>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
>>> min(repeat(lambda: merge_two_dicts(x, y)))
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries


Answer #2

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
{"a": 1, "c": 11, "b": 10}


Answer #3

An alternative:

z = x.copy()

Separation of business logic and data access in django

2 answers

I am writing a project in Django and I see that 80% of the code is in the file This code is confusing and, after a certain time, I cease to understand what is really happening.

Here is what bothers me:

  1. I find it ugly that my model level (which was supposed to be responsible only for the work with data from a database) is also sending email, walking on API to other services, etc.
  2. Also, I find it unacceptable to place business logic in the view, because this way it becomes difficult to control. For example, in my application there are at least three ways to create new instances of User, but technically it should create them uniformly.
  3. I do not always notice when the methods and properties of my models become non-deterministic and when they develop side effects.

Here is a simple example. At first, the User model was like this:

class User(db.Models):

    def get_present_name(self):
        return or "Anonymous"

    def activate(self):
        self.status = "activated"

Over time, it turned into this:

class User(db.Models):

    def get_present_name(self): 
        # property became non-deterministic in terms of database
        # data is taken from another service by api
        return remote_api.request_user_name(self.uid) or "Anonymous" 

    def activate(self):
        # method now has a side effect (send message to user)
        self.status = "activated"
        send_mail("Your account is activated!", "…", [])

What I want is to separate entities in my code:

  1. Entities of my database, persistence level: What data does my application keep?
  2. Entities of my application, business logic level: What does my application do?

What are the good practices to implement such an approach that can be applied in Django?


Answer #1

It seems like you are asking about the difference between the data model and the domain model – the latter is where you can find the business logic and entities as perceived by your end user, the former is where you actually store your data.

Furthermore, I"ve interpreted the 3rd part of your question as: how to notice failure to keep these models separate.

These are two very different concepts and it"s always hard to keep them separate. However, there are some common patterns and tools that can be used for this purpose.

About the Domain Model

The first thing you need to recognize is that your domain model is not really about data; it is about actions and questions such as "activate this user", "deactivate this user", "which users are currently activated?", and "what is this user"s name?". In classical terms: it"s about queries and commands.

Thinking in Commands

Let"s start by looking at the commands in your example: "activate this user" and "deactivate this user". The nice thing about commands is that they can easily be expressed by small given-when-then scenario"s:

given an inactive user
when the admin activates this user
then the user becomes active
and a confirmation e-mail is sent to the user
and an entry is added to the system log
(etc. etc.)

Such scenario"s are useful to see how different parts of your infrastructure can be affected by a single command – in this case your database (some kind of "active" flag), your mail server, your system log, etc.

Such scenario"s also really help you in setting up a Test Driven Development environment.

And finally, thinking in commands really helps you create a task-oriented application. Your users will appreciate this :-)

Expressing Commands

Django provides two easy ways of expressing commands; they are both valid options and it is not unusual to mix the two approaches.

The service layer

The service module has already been described by @Hedde. Here you define a separate module and each command is represented as a function.

def activate_user(user_id):
    user = User.objects.get(pk=user_id)

    # set active flag = True

    # mail user

    # etc etc

Using forms

The other way is to use a Django Form for each command. I prefer this approach, because it combines multiple closely related aspects:

  • execution of the command (what does it do?)
  • validation of the command parameters (can it do this?)
  • presentation of the command (how can I do this?)

class ActivateUserForm(forms.Form):

    user_id = IntegerField(widget = UsernameSelectWidget, verbose_name="Select a user to activate")
    # the username select widget is not a standard Django widget, I just made it up

    def clean_user_id(self):
        user_id = self.cleaned_data["user_id"]
        if User.objects.get(pk=user_id).active:
            raise ValidationError("This user cannot be activated")
        # you can also check authorizations etc. 
        return user_id

    def execute(self):
        This is not a standard method in the forms API; it is intended to replace the 
        "extract-data-from-form-in-view-and-do-stuff" pattern by a more testable pattern. 
        user_id = self.cleaned_data["user_id"]

        user = User.objects.get(pk=user_id)

        # set active flag = True

        # mail user

        # etc etc

Thinking in Queries

You example did not contain any queries, so I took the liberty of making up a few useful queries. I prefer to use the term "question", but queries is the classical terminology. Interesting queries are: "What is the name of this user?", "Can this user log in?", "Show me a list of deactivated users", and "What is the geographical distribution of deactivated users?"

Before embarking on answering these queries, you should always ask yourself this question, is this:

  • a presentational query just for my templates, and/or
  • a business logic query tied to executing my commands, and/or
  • a reporting query.

Presentational queries are merely made to improve the user interface. The answers to business logic queries directly affect the execution of your commands. Reporting queries are merely for analytical purposes and have looser time constraints. These categories are not mutually exclusive.

The other question is: "do I have complete control over the answers?" For example, when querying the user"s name (in this context) we do not have any control over the outcome, because we rely on an external API.

Making Queries

The most basic query in Django is the use of the Manager object:


Of course, this only works if the data is actually represented in your data model. This is not always the case. In those cases, you can consider the options below.

Custom tags and filters

The first alternative is useful for queries that are merely presentational: custom tags and template filters.


<h1>Welcome, {{ user|friendly_name }}</h1>

def friendly_name(user):
    return remote_api.get_cached_name(

Query methods

If your query is not merely presentational, you could add queries to your (if you are using that), or introduce a module:

def inactive_users():
    return User.objects.filter(active=False)

def users_called_publysher():
    for user in User.objects.all():
        if remote_api.get_cached_name( == "publysher":
            yield user 

Proxy models

Proxy models are very useful in the context of business logic and reporting. You basically define an enhanced subset of your model. You can override a Manager’s base QuerySet by overriding the Manager.get_queryset() method.

class InactiveUserManager(models.Manager):
    def get_queryset(self):
        query_set = super(InactiveUserManager, self).get_queryset()
        return query_set.filter(active=False)

class InactiveUser(User):
    >>> for user in InactiveUser.objects.all():
    …        assert is False 

    objects = InactiveUserManager()
    class Meta:
        proxy = True

Query models

For queries that are inherently complex, but are executed quite often, there is the possibility of query models. A query model is a form of denormalization where relevant data for a single query is stored in a separate model. The trick of course is to keep the denormalized model in sync with the primary model. Query models can only be used if changes are entirely under your control.

class InactiveUserDistribution(models.Model):
    country = CharField(max_length=200)
    inactive_user_count = IntegerField(default=0)

The first option is to update these models in your commands. This is very useful if these models are only changed by one or two commands.

class ActivateUserForm(forms.Form):
    # see above
    def execute(self):
        # see above
        query_model = InactiveUserDistribution.objects.get_or_create(
        query_model.inactive_user_count -= 1

A better option would be to use custom signals. These signals are of course emitted by your commands. Signals have the advantage that you can keep multiple query models in sync with your original model. Furthermore, signal processing can be offloaded to background tasks, using Celery or similar frameworks.

user_activated = Signal(providing_args = ["user"])
user_deactivated = Signal(providing_args = ["user"])

class ActivateUserForm(forms.Form):
    # see above
    def execute(self):
        # see above
        user_activated.send_robust(sender=self, user=user)

class InactiveUserDistribution(models.Model):
    # see above

def on_user_activated(sender, **kwargs):
        user = kwargs["user"]
        query_model = InactiveUserDistribution.objects.get_or_create(
        query_model.inactive_user_count -= 1

Keeping it clean

When using this approach, it becomes ridiculously easy to determine if your code stays clean. Just follow these guidelines:

  • Does my model contain methods that do more than managing database state? You should extract a command.
  • Does my model contain properties that do not map to database fields? You should extract a query.
  • Does my model reference infrastructure that is not my database (such as mail)? You should extract a command.

The same goes for views (because views often suffer from the same problem).

  • Does my view actively manage database models? You should extract a command.

Some References

Django documentation: proxy models

Django documentation: signals

Architecture: Domain Driven Design

Haversine Formula in Python (Bearing and Distance between two GPS points)

2 answers


I would like to know how to get the distance and bearing between 2 GPS points. I have researched on the haversine formula. Someone told me that I could also find the bearing using the same data.


Everything is working fine but the bearing doesn"t quite work right yet. The bearing outputs negative but should be between 0 - 360 degrees. The set data should make the horizontal bearing 96.02166666666666 and is:

Start point: 53.32055555555556 , -1.7297222222222221   
Bearing:  96.02166666666666  
Distance: 2 km  
Destination point: 53.31861111111111, -1.6997222222222223  
Final bearing: 96.04555555555555

Here is my new code:

from math import *

Aaltitude = 2000
Oppsite  = 20000

lat1 = 53.32055555555556
lat2 = 53.31861111111111
lon1 = -1.7297222222222221
lon2 = -1.6997222222222223

lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

dlon = lon2 - lon1
dlat = lat2 - lat1
a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
c = 2 * atan2(sqrt(a), sqrt(1-a))
Base = 6371 * c

Bearing =atan2(cos(lat1)*sin(lat2)-sin(lat1)*cos(lat2)*cos(lon2-lon1), sin(lon2-lon1)*cos(lat2)) 

Bearing = degrees(Bearing)
print ""
print ""
print "--------------------"
print "Horizontal Distance:"
print Base
print "--------------------"
print "Bearing:"
print Bearing
print "--------------------"

Base2 = Base * 1000
distance = Base * 2 + Oppsite * 2 / 2
Caltitude = Oppsite - Aaltitude

a = Oppsite/Base
b = atan(a)
c = degrees(b)

distance = distance / 1000

print "The degree of vertical angle is:"
print c
print "--------------------"
print "The distance between the Balloon GPS and the Antenna GPS is:"
print distance
print "--------------------"

Answer #1

Here"s a Python version:

from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):
    Calculate the great circle distance in kilometers between two points 
    on the earth (specified in decimal degrees)
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    r = 6371 # Radius of earth in kilometers. Use 3956 for miles. Determines return value units.
    return c * r

We hope this article has helped you to resolve the problem. Apart from Max retries exceeded with URL in requests, check other sin-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:

Jan Porretti

Berlin | 2022-12-03

StackOverflow is always a bit confusing 😭 Max retries exceeded with URL in requests is not the only problem I encountered. Will use it in my bachelor thesis

Walter Robinson

Warsaw | 2022-12-03

Thanks for explaining! I was stuck with Max retries exceeded with URL in requests for some hours, finally got it done 🤗. Will use it in my bachelor thesis

Anna Sikorski

Vigrinia | 2022-12-03

Maybe there are another answers? What Max retries exceeded with URL in requests exactly means?. Will use it in my bachelor thesis


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