Related
I have a method to validate input:
def validate_user_input(*args):
for item in args:
if not re.match('^[a-zA-Z0-9_-]+$', item):
And I'm calling it like this:
validate_user_input(var1, var2, ..., var7)
But those are generated from user input, and some of those can be missing. What would be the proper way to do that, without creating tons of if statements?
Variables are assigned from a json input like so, and json input might not have some of the needed properties:
var1 = request.json.get('var1')
I assume they are <class 'NoneType'>
Here's the error: TypeError: expected string or buffer
If your request.json object is a dict or dict-like you can just pass a default value as second argument to get
If I understand correctly you are generating var_ variables by request.json.get('var_') which will either return a string which you want to validate or None if the field was missing.
If this is the case then you can just add a special case to validate_user_input for a None value:
def validate_user_input(*args):
for item in args:
if item is None:
continue #this is acceptable, don't do anything with it
elif not re.match('^[a-zA-Z0-9_-]+$', item):
...
Or it may make more sense to store all of the values you are interested in in a dictionary:
wanted_keys = {'var1','var2','var3'}
## set intersection works in python3
present_keys = wanted_keys & response.json.keys()
## or for python 2 use a basic list comp
#present_keys = [key for key in response.json.keys() if key in wanted_keys]
actual_data = {key: response.json[key] for key in present_keys}
Then you would pass actual_data.values() as the argument list to validate_user_input.
If it really is possible that some var-variables are undefined when you call validate_user_input, why not just initialize them all (e.g. to the empty string '' so that your regex fails) before actually defining them?
On Codewars.com I encountered the following task:
Create a function add that adds numbers together when called in succession. So add(1) should return 1, add(1)(2) should return 1+2, ...
While I'm familiar with the basics of Python, I've never encountered a function that is able to be called in such succession, i.e. a function f(x) that can be called as f(x)(y)(z).... Thus far, I'm not even sure how to interpret this notation.
As a mathematician, I'd suspect that f(x)(y) is a function that assigns to every x a function g_{x} and then returns g_{x}(y) and likewise for f(x)(y)(z).
Should this interpretation be correct, Python would allow me to dynamically create functions which seems very interesting to me. I've searched the web for the past hour, but wasn't able to find a lead in the right direction. Since I don't know how this programming concept is called, however, this may not be too surprising.
How do you call this concept and where can I read more about it?
I don't know whether this is function chaining as much as it's callable chaining, but, since functions are callables I guess there's no harm done. Either way, there's two ways I can think of doing this:
Sub-classing int and defining __call__:
The first way would be with a custom int subclass that defines __call__ which returns a new instance of itself with the updated value:
class CustomInt(int):
def __call__(self, v):
return CustomInt(self + v)
Function add can now be defined to return a CustomInt instance, which, as a callable that returns an updated value of itself, can be called in succession:
>>> def add(v):
... return CustomInt(v)
>>> add(1)
1
>>> add(1)(2)
3
>>> add(1)(2)(3)(44) # and so on..
50
In addition, as an int subclass, the returned value retains the __repr__ and __str__ behavior of ints. For more complex operations though, you should define other dunders appropriately.
As #Caridorc noted in a comment, add could also be simply written as:
add = CustomInt
Renaming the class to add instead of CustomInt also works similarly.
Define a closure, requires extra call to yield value:
The only other way I can think of involves a nested function that requires an extra empty argument call in order to return the result. I'm not using nonlocal and opt for attaching attributes to the function objects to make it portable between Pythons:
def add(v):
def _inner_adder(val=None):
"""
if val is None we return _inner_adder.v
else we increment and return ourselves
"""
if val is None:
return _inner_adder.v
_inner_adder.v += val
return _inner_adder
_inner_adder.v = v # save value
return _inner_adder
This continuously returns itself (_inner_adder) which, if a val is supplied, increments it (_inner_adder += val) and if not, returns the value as it is. Like I mentioned, it requires an extra () call in order to return the incremented value:
>>> add(1)(2)()
3
>>> add(1)(2)(3)() # and so on..
6
You can hate me, but here is a one-liner :)
add = lambda v: type("", (int,), {"__call__": lambda self, v: self.__class__(self + v)})(v)
Edit: Ok, how this works? The code is identical to answer of #Jim, but everything happens on a single line.
type can be used to construct new types: type(name, bases, dict) -> a new type. For name we provide empty string, as name is not really needed in this case. For bases (tuple) we provide an (int,), which is identical to inheriting int. dict are the class attributes, where we attach the __call__ lambda.
self.__class__(self + v) is identical to return CustomInt(self + v)
The new type is constructed and returned within the outer lambda.
If you want to define a function to be called multiple times, first you need to return a callable object each time (for example a function) otherwise you have to create your own object by defining a __call__ attribute, in order for it to be callable.
The next point is that you need to preserve all the arguments, which in this case means you might want to use Coroutines or a recursive function. But note that Coroutines are much more optimized/flexible than recursive functions, specially for such tasks.
Here is a sample function using Coroutines, that preserves the latest state of itself. Note that it can't be called multiple times since the return value is an integer which is not callable, but you might think about turning this into your expected object ;-).
def add():
current = yield
while True:
value = yield current
current = value + current
it = add()
next(it)
print(it.send(10))
print(it.send(2))
print(it.send(4))
10
12
16
Simply:
class add(int):
def __call__(self, n):
return add(self + n)
If you are willing to accept an additional () in order to retrieve the result you can use functools.partial:
from functools import partial
def add(*args, result=0):
return partial(add, result=sum(args)+result) if args else result
For example:
>>> add(1)
functools.partial(<function add at 0x7ffbcf3ff430>, result=1)
>>> add(1)(2)
functools.partial(<function add at 0x7ffbcf3ff430>, result=3)
>>> add(1)(2)()
3
This also allows specifying multiple numbers at once:
>>> add(1, 2, 3)(4, 5)(6)()
21
If you want to restrict it to a single number you can do the following:
def add(x=None, *, result=0):
return partial(add, result=x+result) if x is not None else result
If you want add(x)(y)(z) to readily return the result and be further callable then sub-classing int is the way to go.
The pythonic way to do this would be to use dynamic arguments:
def add(*args):
return sum(args)
This is not the answer you're looking for, and you may know this, but I thought I would give it anyway because if someone was wondering about doing this not out of curiosity but for work. They should probably have the "right thing to do" answer.
Here is my problem I have a number of functions defined and I want to loop through a list of these functions and run them one at a time in the correct order.
def one():
print "One "
def two():
print "Two "
def three(): "Three "
print "Three "
arr = ('one','two','three')
for fnc in arr:
<some how run the function name in the variable fnc>
Any Help Appreciated, as I am a beginner with python and django.
Python functions are first order objects; just put them in the sequence:
arr = (one, two, three)
for fnc in arr:
fnc()
You could store strings too, but then you need to turn those back into the function object first. That'd just be extra busywork you don't really need to do.
You can still turn strings into objects; the globals() function gives you the current global namespace as a dictionary, so globals()['one'] gives you the object referenced by the name one, but this would also give you access to every global in your module; if you then made a mistake it could lead to hard to track bugs or even security holes (as end-users could potentially abuse functions you didn't intent to be called).
If you really need to map names to functions, because, say, you need to take input from something else that only produces strings, use a predefined dictionary:
functions = {
'one': one,
'two': two,
'three': three,
}
and map your string to the function:
function_to_call = 'one'
functions[function_to_call]()
Your function names do not need to match the string values here. By using a dedicated dictionary you limit what can be called.
It depends on where the functions are defined, but if they are in the current context, you can get a reference to them by retrieving them from the globals function:
def fn():
return ":)"
for f in['fn']:
print globals()[f]()
Seems to work...
method_name = 'one'
possibles = globals().copy()
possibles.update(locals())
method = possibles.get(method_name)
if not method:
raise Exception("Method %s not implemented" % method_name)
returned_value = method()
For your specific example, simply use eval:
arr = ('one','two','three')
for fnc in arr:
eval(fnc + '()')
Be aware that using eval() is considered bad practice by some.
I have python 3 code that is not working as expected:
def addFunc(x,y):
print (x+y)
def subABC(x,y,z):
print (x-y-z)
def doublePower(base,exp):
print(2*base**exp)
def RootFunc(inputDict):
for k,v in inputDict.items():
if v[0]==1:
d[k] = addFunc(*v[1:])
elif v[0] ==2:
d[k] = subABC(*v[1:])
elif v[0]==3:
d[k] = doublePower(*v[1:])
d={"s1_7":[1,5,2],"d1_6":[2,12,3,3],"e1_3200":[3,40,2],"s2_13":[1,6,7],"d2_30":[2,42,2,10]}
RootFunc(d)
#test to make sure key var assignment works
print(d)
I get:
{'d2_30': None, 's2_13': None, 's1_7': None, 'e1_3200': None, 'd1_6': None}
I expected:
{'d2_30': 30, 's2_13': 13, 's1_7': 7, 'e1_3200': 3200, 'd1_6': 6}
What's wrong?
Semi related: I know dictionaries are unordered but is there any reason why python picked this order? Does it run the keys through a randomizer?
print does not return a value. It returns None, so every time you call your functions, they're printing to standard output and returning None. Try changing all print statements to return like so:
def addFunc(x,y):
return x+y
This will give the value x+y back to whatever called the function.
Another problem with your code (unless you meant to do this) is that you define a dictionary d and then when you define your function, you are working on this dictionary d and not the dictionary that is 'input':
def RootFunc(inputDict):
for k,v in inputDict.items():
if v[0]==1:
d[k] = addFunc(*v[1:])
Are you planning to always change d and not the dictionary that you are iterating over, inputDict?
There may be other issues as well (accepting a variable number of arguments within your functions, for instance), but it's good to address one problem at a time.
Additional Notes on Functions:
Here's some sort-of pseudocode that attempts to convey how functions are often used:
def sample_function(some_data):
modified_data = []
for element in some_data:
do some processing
add processed crap to modified_data
return modified_data
Functions are considered 'black box', which means you structure them so that you can dump some data into them and they always do the same stuff and you can call them over and over again. They will either return values or yield values or update some value or attribute or something (the latter are called 'side effects'). For the moment, just pay attention to the return statement.
Another interesting thing is that functions have 'scope' which means that when I just defined it with a fake-name for the argument, I don't actually have to have a variable called "some_data". I can pass whatever I want to the function, but inside the function I can refer to the fake name and create other variables that really only matter within the context of the function.
Now, if we run my function above, it will go ahead and process the data:
sample_function(my_data_set)
But this is often kind of pointless because the function is supposed to return something and I didn't do anything with what it returned. What I should do is assign the value of the function and its arguments to some container so I can keep the processed information.
my_modified_data = sample_function(my_data_set)
This is a really common way to use functions and you'll probably see it again.
One Simple Way to Approach Your Problem:
Taking all this into consideration, here is one way to solve your problem that comes from a really common programming paradigm:
def RootFunc(inputDict):
temp_dict = {}
for k,v in inputDict.items():
if v[0]==1:
temp_dict[k] = addFunc(*v[1:])
elif v[0] ==2:
temp_dict[k] = subABC(*v[1:])
elif v[0]==3:
temp_dict[k] = doublePower(*v[1:])
return temp_dict
inputDict={"s1_7":[1,5,2],"d1_6":[2,12,3,3],"e1_3200":[3,40,2],"s2_13":[1,6,7],"d2_30"[2,42,2,10]}
final_dict = RootFunc(inputDict)
As erewok stated, you are using "print" and not "return" which may be the source of your error. And as far as the ordering is concerned, you already know that dictionaries are unordered, according to python doc at least, the ordering is not random, but rather implemented as hash tables.
Excerpt from the python doc: [...]A mapping object maps hashable values to arbitrary objects. Mappings are mutable objects. There is currently only one standard mapping type, the dictionary. [...]
Now key here is that the order of the element is not really random. I have often noticed that the order stays the same no matter how I construct a dictionary on some values... using lambda or just creating it outright, the order has always remained the same, so it can't be random, but it's definitely arbitrary.
This question already has answers here:
Replacements for switch statement in Python?
(44 answers)
Closed 27 days ago.
I have tried making a switch like statement in python, instead of having a lot of if statements.
The code looks like this:
def findStuff(cds):
L=[]
c=0
for i in range(0, len(cds), 3):
a=differencesTo(cds[i:i+3])
result = {
a[2][0]==1: c=i+1,
a[2][1]==1: c=i+2,
a[2][2]==1: c=i+3,
a[1]==1: L.append((cds[i:i+3], a[0], c))
}
return L
My problem is, that this does not work. (Works with if statements, but this would in my opinion be more pretty).
I have found some examples of switches in Python, and they follow this structure. Can anyone help me?
(a) I fail to see what is wrong with if...elif...else
(b) I assume that python does not have a switch statement for the same reason that Smalltalk doesn't: it's almost completely redundant, and in the case where you want to switch on types, you can add an appropriate method to your classes; and likewise switching on values should be largely redundant.
Note: I am informed in the comments that whatever Guido's reason for not creating a switch in the first place, PEPs to have it added were rejected on the basis that support for adding such a statement is extremely limited. See: http://www.python.org/dev/peps/pep-3103/
(c) If you really need switching behaviour, use a hashtable (dict) to store callables. The structure is:
switch_dict = {
Foo: self.doFoo,
Bar: self.doBar,
}
func = switch_dict[switch_var]
result = func() # or if they take args, pass args
There's nothing wrong with a long if:
if switch == 'case0':
do_case0()
elif switch == 'case1':
do_case1()
elif switch == 'case2':
do_case2()
...
If that's too long winded, or if you have a lot of cases, put them in a dictionary:
switch = {'case0': do_case0, 'case1': do_case1, 'case2': do_case2, ...}
switch[case_variable]()
// Alternative:
(switch[case_variable]).__call__()
If your conditions are a bit more complex, you need to think a little about your data structures. e.g.:
switch = {
(0,21): 'never have a pension',
(21,50): 'might have a pension',
(50,65): 'definitely have a pension',
(65, 200): 'already collecting pension'
}
for key, value in switch.items():
if key[0] <= case_var < key[1]:
print(value)
Other ans are suitable for older version of python. For python v3.10+ you can use match/case which is more powerful than general switch/case construct.
def something(val):
match val:
case "A":
return "A"
case "B":
return "B"
case "C":
return "C"
case _:
return "Default"
something("A")
Assignment in Python is a statement, and cannot be a part of expression. Also, using literal in this way evaluates everything at once, which is probably not what you want. Just use ifs, you won't gain any readability by using this.
I don't know which article you've found to do something like this, but this is really messy: the whole result diction will be always evaluated, and instead of doing only part of the work (as a switch / if do), you'll do the whole work everytime. (even if you use only a part of the result).
Really, a fast switch statement in Python is using "if":
if case == 1:
pass
elif case == 2:
pass
elif case == 3:
pass
else:
# default case
pass
With "get" method, you can have the same effect as "switch..case" in C.
Marcin example :
switch_dict = {
Foo: self.doFoo,
Bar: self.doBar,
}
func = switch_dict.get(switch_var, self.dodefault)
result = func() # or if they take args, pass args
You can do something like what you want, but you shouldn't. That said, here's how; you can see how it does not improve things.
The biggest problem with the way you have it is that Python will evaluate your tests and results once, at the time you declare the dictionary. What you'd have to do instead is make all conditions and the resulting statements functions; this way, evaluation is deferred until you call them. Fortunately there is a way to do this inline for simple functions using the lambda keyword. Secondly, the assignment statement can't be used as a value in Python, so our action functions (which are executed if the corresponding condition function returns a truthy value) have to return a value that will be used to increment c; they can't assign to c themselves.
Also, the items in a dictionary aren't ordered, so your tests won't necessarily be performed in the order you define them, meaning you probably should use something other than a dictionary that preserves order, such as a tuple or a list. I am assuming you want only ever one case to execute.
So, here we go:
def findStuff(cds):
cases = [ (lambda: a[2][0] == 1, lambda: i + 1),
(lambda: a[2][1] == 1, lambda: i + 2),
(lambda: a[2][2] == 1, lambda: i + 3),
(lambda: a[1] == 1, lambda: L.append(cds[i:i+3], a[0], c) or 0)
]
L=[]
c=0
for i in range(0, len(cds), 3):
a=differencesTo(cds[i:i+3])
for condition, action in cases:
if condition():
c += action()
break
return L
Is this more readable than a sequence of if/elif statements? Nooooooooooooo. In particular, the fourth case is far less comprehensible than it should be because we are having to rely on a function that returns the increment for c to modify a completely different variable, and then we have to figure out how to get it to return a 0 so that c won't actually be modified. Uuuuuugly.
Don't do this. In fact this code probably won't even run as-is, as I deemed it too ugly to test.
While there is nothing wrong with if..else, I find "switch in Python" still an intriguing problem statement. On that, I think Marcin's (deprecated) option (c) and/or Snim2's second variant can be written in a more readable way.
For this we can declare a switch class, and exploit the __init__() to declare the case we want to switch, while __call__() helps to hand over a dict listing the (case, function) pairs:
class switch(object):
def __init__(self, case):
self._case = case
def __call__(self, dict_):
try:
return dict_[self._case]()
except KeyError:
if 'else' in dict_:
return dict_['else']()
raise Exception('Given case wasn\'t found.')
Or, respectively, since a class with only two methods, of which one is __init__(), isn't really a class:
def switch(case):
def cases(dict_):
try:
return dict_[case]()
except KeyError:
if 'else' in dict_:
return dict_['else']()
raise Exception('Given case wasn\'t found.')
return cases
(note: choose something smarter than Exception)
With for example
def case_a():
print('hello world')
def case_b():
print('sth other than hello')
def default():
print('last resort')
you can call
switch('c') ({
'a': case_a,
'b': case_b,
'else': default
})
which, for this particular example would print
last resort
This doesn't behave like a C switch in that there is no break for the different cases, because each case executes only the function declared for the particular case (i.e. break is implicitly always called). Secondly, each case can list exactly only one function that will be executed upon a found case.