How to use pytest_collection method in Pytest

try:
    thevariable
except NameError:
    print("well, it WASN'T defined after all!")
else:
    print("sure, it was defined.")

Either downgrade to pytest-bdd<4 where this behaviour is still accepted, or rename the steps by removing the test_ prefix to prevent pytest from recognizing them as separate tests.

@when(...)
def register(admin_login, ...):
    ...

@then(...)
def login(admin_login):
    ...

should work.

Putting hpaulj's answer into actual code, something like this works:

class CustomHelpFormatter(argparse.HelpFormatter):
    def _format_action_invocation(self, action):
        if not action.option_strings or action.nargs == 0:
            return super()._format_action_invocation(action)
        default = self._get_default_metavar_for_optional(action)
        args_string = self._format_args(action, default)
        return ', '.join(action.option_strings) + ' ' + args_string

fmt = lambda prog: CustomHelpFormatter(prog)
parser = argparse.ArgumentParser(formatter_class=fmt)

To additionally extend the default column size for help variables, add constructor to CustomHelpFormatter:

def __init__(self, prog):
    super().__init__(prog, max_help_position=40, width=80)

Seeing it in action:

usage: bk set [-h] [-p] [-s r] [-f] [-c] [-b c] [-t x y] [-bs s] [-bc c]
              [--crop x1 y1 x2 y2] [-g u r d l]
              monitor [path]

positional arguments:
  monitor                    monitor number
  path                       input image path

optional arguments:
  -h, --help                 show this help message and exit
  -p, --preview              previews the changes without applying them
  -s, --scale r              scales image by given factor
  -f, --fit                  fits the image within monitor work area
  -c, --cover                makes the image cover whole monitor work area
  -b, --background c         selects background color
  -t, --translate x y        places the image at given position
  -bs, --border-size s       selects border width
  -bc, --border-color c      selects border size
  --crop x1 y1 x2 y2         selects crop area
  -g, --gap, --gaps u r d l  keeps "border" around work area

Turning my comment into an answer, since pytest==6.0 was released yesterday:

This was fixed in version 6.0 with an introduction of a new parser for -k argument. The old behaviour of evaling -k argument as an expression was removed. Anyone interested in the actual implementation should take a look at #7122's diff.

I think what you are looking for is nose. It works like a test runner for unittest.

You can drop into the debugger on errors, with the following command:

nosetests --pdb

Code

f'{path}/{filename}.db' 

creates string.

You should use

path / f'{filename}.db'

to get Path object.

And then you can do

file = db_helper('testfile')
file.unlink()

---

BTW:

You can also use

    home = Path.home()
    path = home / '.local/apikeychain'

or shorter

    path = Path.home() / '.local/apikeychain'

to make code simpler.

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

For dictionaries x and y, their shallowly-merged dictionary z takes values from y, replacing those from x.

• In Python 3.9.0 or greater (released 17 October 2020, PEP-584, discussed here):

  z = x | y
  

• 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)
  

Explanation

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()
    z.update(y)
    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:
        result.update(dictionary)
    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.

and

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

Usage:

>>> 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()
    z.update(y)
    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}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954

$ 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

• My explanation of Python's dictionary implementation, updated for 3.6.
• Answer on how to add new keys to a dictionary
• Mapping two lists into a dictionary
• The official Python docs on dictionaries
• The Dictionary Even Mightier - talk by Brandon Rhodes at Pycon 2017
• Modern Python Dictionaries, A Confluence of Great Ideas - talk by Raymond Hettinger at Pycon 2017

Since you have a GraphQL server, all you need is a GraphQL client implementation. From official documentation, you can find the latest implementations of Python GraphQL client libraries according to your python version needs. Find more here

For you could use the GQL package from https://github.com/graphql-python/gql like so,

from gql import gql, Client

client = Client(schema=schema)
query = gql('''
{
  hello
}
''')

client.execute(query)

Keep in mind, that since GraphQL is a specification you could also write a Client of your own. The GraphQL specification isn’t opinionated about the transportation layer, so it can be any. An example of HTTP implementation of it can be found at https://www.robinwieruch.de/react-graphql-client-library/

Quick Answer:

The simplest way to get row counts per group is by calling .size(), which returns a Series:

df.groupby(['col1','col2']).size()

Usually you want this result as a DataFrame (instead of a Series) so you can do:

df.groupby(['col1', 'col2']).size().reset_index(name='counts')

If you want to find out how to calculate the row counts and other statistics for each group continue reading below.

---

Detailed example:

Consider the following example dataframe:

In [2]: df
Out[2]: 
  col1 col2  col3  col4  col5  col6
0    A    B  0.20 -0.61 -0.49  1.49
1    A    B -1.53 -1.01 -0.39  1.82
2    A    B -0.44  0.27  0.72  0.11
3    A    B  0.28 -1.32  0.38  0.18
4    C    D  0.12  0.59  0.81  0.66
5    C    D -0.13 -1.65 -1.64  0.50
6    C    D -1.42 -0.11 -0.18 -0.44
7    E    F -0.00  1.42 -0.26  1.17
8    E    F  0.91 -0.47  1.35 -0.34
9    G    H  1.48 -0.63 -1.14  0.17

First let's use .size() to get the row counts:

In [3]: df.groupby(['col1', 'col2']).size()
Out[3]: 
col1  col2
A     B       4
C     D       3
E     F       2
G     H       1
dtype: int64

Then let's use .size().reset_index(name='counts') to get the row counts:

In [4]: df.groupby(['col1', 'col2']).size().reset_index(name='counts')
Out[4]: 
  col1 col2  counts
0    A    B       4
1    C    D       3
2    E    F       2
3    G    H       1

Including results for more statistics

When you want to calculate statistics on grouped data, it usually looks like this:

In [5]: (df
   ...: .groupby(['col1', 'col2'])
   ...: .agg({
   ...:     'col3': ['mean', 'count'], 
   ...:     'col4': ['median', 'min', 'count']
   ...: }))
Out[5]: 
            col4                  col3      
          median   min count      mean count
col1 col2                                   
A    B    -0.810 -1.32     4 -0.372500     4
C    D    -0.110 -1.65     3 -0.476667     3
E    F     0.475 -0.47     2  0.455000     2
G    H    -0.630 -0.63     1  1.480000     1

The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.

To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:

In [6]: gb = df.groupby(['col1', 'col2'])
   ...: counts = gb.size().to_frame(name='counts')
   ...: (counts
   ...:  .join(gb.agg({'col3': 'mean'}).rename(columns={'col3': 'col3_mean'}))
   ...:  .join(gb.agg({'col4': 'median'}).rename(columns={'col4': 'col4_median'}))
   ...:  .join(gb.agg({'col4': 'min'}).rename(columns={'col4': 'col4_min'}))
   ...:  .reset_index()
   ...: )
   ...: 
Out[6]: 
  col1 col2  counts  col3_mean  col4_median  col4_min
0    A    B       4  -0.372500       -0.810     -1.32
1    C    D       3  -0.476667       -0.110     -1.65
2    E    F       2   0.455000        0.475     -0.47
3    G    H       1   1.480000       -0.630     -0.63

---

Footnotes

The code used to generate the test data is shown below:

In [1]: import numpy as np
   ...: import pandas as pd 
   ...: 
   ...: keys = np.array([
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['A', 'B'],
   ...:         ['C', 'D'],
   ...:         ['C', 'D'],
   ...:         ['C', 'D'],
   ...:         ['E', 'F'],
   ...:         ['E', 'F'],
   ...:         ['G', 'H'] 
   ...:         ])
   ...: 
   ...: df = pd.DataFrame(
   ...:     np.hstack([keys,np.random.randn(10,4).round(2)]), 
   ...:     columns = ['col1', 'col2', 'col3', 'col4', 'col5', 'col6']
   ...: )
   ...: 
   ...: df[['col3', 'col4', 'col5', 'col6']] = \
   ...:     df[['col3', 'col4', 'col5', 'col6']].astype(float)
   ...: 

Disclaimer:

If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.