I believe this is a simple question but still want to get a quick and clear answer to my case:
def get_query_history(idx, url, archive_location):
idx = idx + 1
return idx # I meant to return the idx's value (end up 1000 for every call) and used it in the next loop in main
main:
idx = 1
while current <= end_date:
with open(archive_location, 'a') as the_archive:
get_query_history(idx, url, archive_location) # I want to increase the idx every time I call the function
Apparently this is not the way I should take in python, can anyone enlighten me?
Here, I'll post it as an answer but I'll expand a bit.
Since you're returning idx increased value, just store it back in the 'main' scope:
idx = 1
while current <= end_date:
with open(archive_location, 'a') as the_archive:
idx = get_query_history(idx, url, archive_location)
# make sure you update your `current` ;)
In some languages you have an option to pass a variable to a function by reference in such a way that the function can change its value so you wouldn't need to return your value. Python kind of passes by reference, but since simple values are unmutable whenever you try to set its value in your function the reference to the passed value gets overwritten.
This doesn't apply to encapsulated objects, tho, so you could encapsulate your idx in a list and then pass it as a list. In that case you wouldn't need return at all:
def get_query_history(idx, url, archive_location):
idx[0] += 1
# do whatever else
# in your main:
idx = [1] # encapsulate the value in a list
while current <= end_date:
with open(archive_location, 'a') as the_archive:
get_query_history(idx, url, archive_location) # notice, no return capture
# make sure you update your `current` ;)
But generally, if you can return the value there is no need for these shenanigans, it's just to demonstrate that a function can modify the passed arguments under certain conditions.
And, finally, if you really want to force pass-by-reference behavior, you can totally hack Python to do even that, check this (and never use it in production!) ;)
Related
So I want to create a function that generates consecutive numbers from 'start' to 'end' as many as 'size'. For the iteration, it will be calculated inside the function. But I have problem to set default value of parameter 'end'. Before I explain further, here's the code:
# Look at this -------------------------------
# ||
# \/
def consecutive_generator(size=20, start=0, end=(size+start)):
i = start
iteration = (end-start)/size
arr = []
temp_size = 0
while temp_size < size:
arr.append(i)
i += iteration
temp_size += 1
return arr
# with default end, so the 'end' parameter will be 11
c1= consecutive_generator(10, start=1)
print(c1)
# with end set
c2= consecutive_generator(10, end=20)
print(c2)
As can be seen above (on the default value of the 'end' parameter), what I want to achieve is the 'end' parameter whose default value is 'start' + 'size' parameters (then the iteration will be 1)
The output will definitely be an error. So how can i do this? (this is my first time asking on stackoverflow sorry if i made a mistake)
(Closed)
This is a pretty standard pattern:
def consecutive_generator(size=20, start=0, end=None):
if end is None:
end = size + start
The default way is as Samwise said but there is an alternative solution that might work just as well.
def consecutive_generator(size=20, start=0, **kwargs):
end = kwargs.get('end', size+start)
This method allows you to get end if it exists or simply set the value of end if it doesn't.
To call it, this method does require that the function call have the parameter end specified if you want to set it to something other than the default.
consecutive_generator(20, 0, end=50)
dict.get
And perhaps consider checking out the documentation for range and numpy.linspace
According to Python's document,
Default parameter values are evaluated from left to right when the function definition is executed.
So default values of parameters are can't evaluated from other dynamic values which are passed when the function called.
You should use the distinguishable value like None for the default value. Then you can check it and evaluate proper default value dynamically. For example,
def consecutive_generator(size=20, start=0, end=None):
if end is None:
end = size + start
...
If you need None as a valid value passed from caller, you can use other object or something to distinguish from valid value.
default_end = object()
def consecutive_generator(size=20, start=0, end=default_end):
if end is default_end:
end = size + start
...
I am new to python and i have this project I am working on a small project with two functions where the first returns the index of the first time a difference is spotted in a string. The next function does that but in a list of strings. Now, due to my being an amateur, i have used an excessive amount of if and else statements which resulted in too many return statements especially in the second function, and i get the error [R1710: inconsistent-return-statements]. How do i fix it and can anybody give me clear examples to better pieces of code? Sorry for the question being so long.
IDENTICAL = -1
def singleline_diff(line1, line2):
"""
Inputs:
line1 - first single line string
line2 - second single line string
Output:
Returns the index where the first difference between
line1 and line2 occurs.
Returns IDENTICAL if the two lines are the same.
"""
len1 = len(line1)
len2 = len(line2)
minimum_length = min(len1, len2)
if len1 != len2:
if minimum_length == 0:
return 0
for idx in range(minimum_length):
if line1[idx] == line2[idx]:
pass
else:
return idx
return idx + 1
for idx in range(len1):
if line1[idx] == line2[idx]:
pass
else:
return idx
return IDENTICAL
def multiline_diff(lines1, lines2):
"""
Inputs:
lines1 - list of single line strings
lines2 - list of single line strings
Output:
Returns a tuple containing the line number (starting from 0) and
the index in that line where the first difference between lines1
and lines2 occurs.
Returns (IDENTICAL, IDENTICAL) if the two lists are the same.
"""
line_no = singleline_diff(lines1, lines2)
len_lines1, len_lines2 = len(lines1), len(lines2)
if len_lines1 == len_lines2:
if (len_lines1 or len_lines2) == 0:
if len_lines1 == len_lines2:
return (IDENTICAL, IDENTICAL)
else:
idx = singleline_diff(lines1[line_no], lines2[line_no])
return (line_no, idx)
else:
idx = singleline_diff(lines1[line_no], lines2[line_no])
if line_no == IDENTICAL:
return (IDENTICAL, IDENTICAL)
elif line_no != IDENTICAL:
return (line_no, idx)
else:
return (line_no, 0)
Where was a semantic mistake in OP's code is in Abhishek Arya's answer
TL;DR - early return:
def your_function():
if not should_do():
return # NO RETURN VALUE!
# rest of the function
...yes, this will no longer emit the inconsistent-return-statements ;)
This Q/A pops also when you search for inconsistent-return-statements, I want to give a brief "common problems" guide for those.
Case A: return value is irrelevant, you just want to exit function early
There are cases, where there are functions (or "procedures" if you want to get technical about it) that just do something, but are not expected to have any return values AT ALL,
at the same time, there may be e.g. some sort of check at the start of the function whether this function run even makes sense, what may first come to your mind, is wrapping the whole function code in an if statement:
def your_function(article):
if display_content():
content = get_content(article)
# do some extensive logic to generate final content
# ...
print(content)
...this is oversimplified, but let's hope you can imagine how such coding can pretty quickly fall into a "spaghetti code" if there are more checks and more code in general + it also steals that one "tab" of a space that you so desperately need to fit into your project's max line length.
Luckily, same as in many other programming languages, there IS a way of an early ending of a function by returning at ANY place within the function run, meaning in any "Control Flow" - including if/elif/else, for/while loops, ...
Now you'd probably jump quick to just return None, False, etc. although it would work, you'd still get the pylint inconsistent-return-statements warning - to understand why let's see the warning's message:
Either all return statements in a function should return an
expression, or none of them should. pylint(inconsistent-return-statements)
From pylint's point of view, if you put anything after the return it will be considered as an expression. So what to do? Actually, in Python, you CAN return "nothing" (again this is not unique to Python)
def your_function(article):
if not display_content():
return
content = get_content(article)
# do some extensive logic to generate final content
# ...
print(content)
Although in Python returning "nothing" should be (and technically, to my knowledge, it is) an equivalent of return None, by physically writing "None" you are expressing the intention no matter the implicity of it.
Don't confuse this though with pylint(assignment-from-none) (Assigning result of a function call, where the function returns None) - where both "return" AND "return None" are considered as returning None!
Case B: Your function has a case when it doesn't return
Quite common mistake especially in a larger code is to create a code part which results in simply not returning anything. This is not exactly OP's case, since they used just a negation of the same condition, but pylint doesn't know that, so here's its thought process:
if SOME_CONDITION: # ok, here's just another condition
return someReturnExpression # and ok, it returns SOMETHING, let's note that
elif OPPOSITE_OF_SOME_CONDITION: # ok, here's just another condition
return someReturnExpression # and ok, it returns SOMETHING, let's note that
# WAIT ! What?! THERE WAS NO "else:"! Hmmm...
# ...what happens if both conditions fail? NOTHING WOULD BE RETURNED!
# We need to make a warning about that!
# (fact that sometimes they return SOMETHING and sometimes NOTHING)
So this inconsistent-return-statements could be resolved with
if SOME_CONDITION: # ok, here's some condition
return someReturnExpression # and ok, it returns SOMETHING, let's note that
else: # ok, here's else
return someReturnExpression # and ok, it returns SOMETHING, let's note that
# Ok, so if returns SOMETHING, else returns SOMETHING,
# so SOMETHING is returned every time! that's good!
...this in itself works, but it will generate yet another pylint issue
Unnecessary "else" after "return" pylint(no-else-return)
See python actually encourages early returns since it often leads to a cleaner code.
return during function run ENDS(/exits) the function and pylint sees that - it sees that if the condition was true, the function code would simply end there - so what it, Abhishek Arya, me and many others suggest is simply continuing with the code after the if part:
if SOME_CONDITION:
return someReturnExpression
# ... some more code ...
# ... some more code ...
return someReturnExpression
Case C: Combination
Simply don't combine "just" return with return SOMETHING,
if you really need to return None, simply explicitly return None in that case
def get_article(id):
article = find_article(id)
if article.id == 0:
return None
return article
This is just an example, this is not how you'd really check for some articles ;)
Look at the code here:
if len_lines1 == len_lines2:
return (IDENTICAL, IDENTICAL)
else:
idx = singleline_diff(lines1[line_no], lines2[line_no])
return (line_no, idx)
You could have written the above thing like:
if len_lines1 == len_lines2:
return (IDENTICAL, IDENTICAL)
idx = singleline_diff(lines1[line_no], lines2[line_no])
return (line_no, idx)
You just don't need an else block to return this expression as this part of code will automatically be called if the control doesn't go into if block. Hope it helps.
I am brushing up a bit of good old algorithms, and doing it with python, since I use it more often nowadays.
I am facing an issue when running a recursive function; where the variable get reset every time that the recursive function call itself:
def recursive_me(mystring):
chars = len(mystring)
if chars is 0:
print("Done")
else:
first = int(str[0])
total = + first
print(total)
recursive_me(mystring[1:])
recursive_me("4567")
What I am doing here is to get a string made of digits; take the first, convert it to an int; and run recursively the function again, so I can take one digit at time from the string and sum all the values.
Ideally the output should show the total, while it add all the digits (4+5+6+7), although when the recursive function is called the first time, the function reset the total value.
Is common habit to use global variables when running operations with recursive functions or am I doing something wrong?
You can code as simply as this:
def recursive_me(mystring):
if mystring: # recursive case
return int(mystring[0]) + recursive_me(mystring[1:])
else: # base case
return 0
or
def recursive_me(mystring, total = 0):
if mystring: # recursive case
return recursive_me(mystring[1:], total + int(mystring[0]))
else: # base case
return total
although this won't help much in Python since it doesn't implement tail-call optimisation.
If you want to see the intermediate values, change the second version like so:
def recursive_me(mystring, total = 0):
if mystring: # recursive case
newtotal = total + int(mystring[0])
print(newtotal)
return recursive_me(mystring[1:], newtotal)
else: # base case
return total
then
4
9
15
22
22 # this is the return value; previous output is from `print()`
as a foreword: a lot of answers received meaningful edits in the meantime I was writing this answer. Don't hold it against me.
I'm throwing my two cents in here just because there's a lot of over-complicated answers.
This is a corrected copy-paste of the OP's effort.
def recursive_me(mystring, total=0):
chars = len(mystring)
if chars is 0:
print("Done")
return total
else:
first = int(mystring[0])
total += first
print(total)
recursive_me(mystring[1:], total)
first what happens is that we check the base case, if there's no left chars in the string. If the string length is 0 we return the total calculated ammount.
Otherwise, we turn the first of the chars into an int, and add it to total. The first error you have is that you wrote str[0]. str is a python built in type and the produced error would be something like "str is not subscriptable".
This error means that the str can't be operated on by "[]" operator. The same would happen if you tried doing 1[0] because 1 is a integer. The "[]" operator can only operate on lists, tuples and strings (I might've forgot some built-in type).
The second error you had was with the addition part. You had written total = + first but the operator you are looking for is the += which in fact is just a shortened way to write a = a+b.
Additionally, your original question was concerning about "python" forgetting the value of "total". This is because you have to either pass that value forward, or write your recursive function in a way that "forces" it to, what's called, evaluate your next call to your function on the spot.
In my example I'm sending the next call of the function recursive_me, the current total value. In the example given by #uselpa; above he's making python evaluate the next call to the function by putting it after operator +:
return int(mystring[0]) + recursive_me(mystring[1:])
this then gets to be (for recursive_me("4567"))
return int(4)+recursive_me("567")
return int(4)+int(5)+recursive_me("67")
....
return int(4)+int(5)+int(6)+int(7)+0
because python needs to return a value here, but the expression keeps calling new functions and python can't return until it evaluates all of them to a final number (in this case at least).
The common practice is to save these variables as parameters, and pass them along the chain. It seems in your case, you would want to pass total as an additional parameter, and update it as needed.
There's also a neat functional way to do it in python
t=raw_input()
print reduce(lambda a, b: a+b, map(int,t))
This is recursive in nature.
Some pointers:
Your default case should return an actual number (0 in your case) and not just print done.
total = + first is setting total to first, not adding first to total. You would need total += first to do the latter.
The trick with "retaining" the value of your current total is to "save" it in the recursive call-chain itself by passing it along with each call. You won't need a global variable or a default parameter to do this.
Here's a solution:
def recursive_me(mystring):
if not mystring: # True if mystring is empty
return 0
return int(mystring[0]) + recursive_me(mystring[1:])
print(recursive_me("4567")) # 22
Here is a solution that uses the LEGB scope rule to avoid creating a new string instance on every recursive call
def sum_str(mystring):
def recursive_me(pos):
cur_char = int(mystring[pos])
if pos:
return cur_char + recursive_me(pos-1)
else:
return cur_char
return recursive_me(len(mystring)-1)
s = '4567'
print('summing', s)
print(sum_str(s))
However, indexing can be avoided as well by iterating on the string
def sum_str(mystring):
def recursive_me(itx):
try:
cur_char = int(next(itx))
return cur_char + recursive_me(itx)
except StopIteration:
return 0
return recursive_me(iter(mystring))
Obviously, both solutions produce
summing 4567
22
I need some help understanding a function that i want to use but I'm not entirely sure what some parts of it do. I understand that the function is creating dictionaries from reads out of a Fasta-file. From what I understand this is supposed to generate pre- and suffix dictionaries for ultimately extending contigs (overlapping dna-sequences).
The code:
def makeSuffixDict(reads, lenSuffix = 20, verbose = True):
lenKeys = len(reads[0]) - lenSuffix
dict = {}
multipleKeys = []
i = 1
for read in reads:
if read[0:lenKeys] in dict:
multipleKeys.append(read[0:lenKeys])
else:
dict[read[0:lenKeys]] = read[lenKeys:]
if verbose:
print("\rChecking suffix", i, "of", len(reads), end = "", flush = True)
i += 1
for key in set(multipleKeys):
del(dict[key])
if verbose:
print("\nCreated", len(dict), "suffixes with length", lenSuffix, \
"from", len(reads), "Reads. (", len(reads) - len(dict), \
"unambigous)")
return(dict)
Additional Information: reads = readFasta("smallReads.fna", verbose = True)
This is how the function is called:
if __name__ == "__main__":
reads = readFasta("smallReads.fna", verbose = True)
suffixDicts = makeSuffixDicts(reads, 10)
The smallReads.fna file contains strings of bases (Dna):
"> read 1
TTATGAATATTACGCAATGGACGTCCAAGGTACAGCGTATTTGTACGCTA
"> read 2
AACTGCTATCTTTCTTGTCCACTCGAAAATCCATAACGTAGCCCATAACG
"> read 3
TCAGTTATCCTATATACTGGATCCCGACTTTAATCGGCGTCGGAATTACT
Here are the parts I don't understand:
lenKeys = len(reads[0]) - lenSuffix
What does the value [0] mean? From what I understand "len" returns the number of elements in a list.
Why is "reads" automatically a list? edit: It seems a Fasta-file can be declared as a List. Can anybody confirm that?
if read[0:lenKeys] in dict:
Does this mean "from 0 to 'lenKeys'"? Still confused about the value.
In another function there is a similar line: if read[-lenKeys:] in dict:
What does the "-" do?
def makeSuffixDict(reads, lenSuffix = 20, verbose = True):
Here I don't understand the parameters: How can reads be a parameter? What is lenSuffix = 20 in the context of this function other than a value subtracted from len(reads[0])?
What is verbose? I have read about a "verbose-mode" ignoring whitespaces but i have never seen it used as a parameter and later as a variable.
The tone of your question makes me feel like you're confusing things like program features (len, functions, etc) with things that were defined by the original programmer (the type of reads, verbose, etc).
def some_function(these, are, arbitrary, parameters):
pass
This function defines a bunch of parameters. They don't mean anything at all, other than the value I give to them implicitly. For example if I do:
def reverse_string(s):
pass
s is probably a string, right? In your example we have:
def makeSuffixDict(reads, lenSuffix = 20, verbose = True):
lenKeys = len(reads[0]) - lenSuffix
...
From these two lines we can infer a few things:
the function will probably return a dictionary (from its name)
lenSuffix is an int, and verbose is a bool (from their default parameters)
reads can be indexed (string? list? tuple?)
the items inside reads have length (string? list? tuple?)
Since Python is dynamically typed, this is ALL WE CAN KNOW about the function so far. The rest would be explained by its documentation or the way it's called.
That said: let me cover all your questions in order:
What does the value [0] mean?
some_object[0] is grabbing the first item in a container. [1,2,3][0] == 1, "Hello, World!"[0] == "H". This is called indexing, and is governed by the __getitem__ magic method
From what I understand "len" returns the number of elements in a list.
len is a built-in function that returns the length of an object. It is governed by the __len__ magic method. len('abc') == 3, also len([1, 2, 3]) == 3. Note that len(['abc']) == 1, since it is measuring the length of the list, not the string inside it.
Why is "reads" automatically a list?
reads is a parameter. It is whatever the calling scope passes to it. It does appear that it expects a list, but that's not a hard and fast rule!
(various questions about slicing)
Slicing is doing some_container[start_idx : end_idx [ : step_size]]. It does pretty much what you'd expect: "0123456"[0:3] == "012". Slice indexes are considered to be zero-indexed and lay between the elements, so [0:1] is identical to [0], except that slices return lists, not individual objects (so 'abc'[0] == 'a' but 'abc'[0:1] == ['a']). If you omit either start or end index, it is treated as the beginning or end of the string respectively. I won't go into step size here.
Negative indexes count from the back, so '0123456'[-3:] == '456'. Note that [-0]is not the last value,[-1]is. This is contrasted with[0]` being the first value.
How can reads be a parameter?
Because the function is defined as makeSuffixDict(reads, ...). That's what a parameter is.
What is lenSuffix = 20 in the context of this function
Looks like it's the length of the expected suffix!
What is verbose?
verbose has no meaning on its own. It's just another parameter. Looks like the author included the verbose flag so you could get output while the function ran. Notice all the if verbose blocks seem to do nothing, just provide feedback to the user.
I'm using a recursive function to sort a list in Python, and I want to keep track of the number of sorts/merges as the function continues. However, when I declare/initialize the variable inside the function, it becomes a local variable inside each successive call of the function. If I declare the variable outside the function, the function thinks it doesn't exist (i.e. has no access to it). How can I share this value across different calls of the function?
I tried to use the "global" variable tag inside and outside the function like this:
global invcount ## I tried here, with and without the global tag
def inv_sort (listIn):
global invcount ## and here, with and without the global tag
if (invcount == undefined): ## can't figure this part out
invcount = 0
#do stuff
But I cannot figure out how to check for the undefined status of the global variable and give it a value on the first recursion call (because on all successive recursions it should have a value and be defined).
My first thought was to return the variable out of each call of the function, but I can't figure out how to pass two objects out of the function, and I already have to pass the list out for the recursion sort to work. My second attempt to resolve this issue involved me adding the variable invcount to the list I'm passing as the last element with an identifier, like "i27". Then I could just check for the presence of the identifier (the letter i in this example) in the last element and if present pop() it off at the beginning of the function call and re-add it during the recursion. In practice this is becoming really convoluted and while it may work eventually, I'm wondering if there is a more practical or easier solution.
Is there a way to share a variable without directly passing/returning it?
There's couple of things you can do. Taking your example you should modify it like this:
invcount = 0
def inv_sort (listIn):
global invcount
invcount += 1
# do stuff
But this approach means that you should zero invcount before each call to inv_sort.
So actually its better to return invcount as a part of result. For example using tuples like this:
def inv_sort(listIn):
#somewhere in your code recursive call
recursive_result, recursive_invcount = inv_sort(argument)
# this_call_invcount includes recursive_invcount
return this_call_result, this_call_invcount
There's no such thing as an "undefined" variable in Python, and you don't need one.
Outside the function, set the variable to 0. Inside the loop, use the global keyword, then increment.
invcount = 0
def inv_sort (listIn):
global invcount
... do stuff ...
invcount += 1
An alternative might be using a default argument, e.g.:
def inv_sort(listIn, invcount=0):
...
invcount += 1
...
listIn, invcount = inv_sort(listIn, invcount)
...
return listIn, invcount
The downside of this is that your calls get slightly less neat:
l, _ = inv_sort(l) # i.e. ignore the second returned parameter
But this does mean that invcount automatically gets reset each time the function is called with a single argument (and also provides the opportunity to inject a value of invcount if necessary for testing: assert result, 6 == inv_sort(test, 5)).
Assuming that you don't need to know the count inside the function, I would approach this using a decorator function:
import functools
def count_calls(f):
#functools.wraps(f)
def func(*args):
func.count += 1
return f(*args)
func.count = 0
return func
You can now decorate your recursive function:
#count_calls
def inv_sort(...):
...
And check or reset the count before or after calling it:
inv_sort.count = 0
l = inv_sort(l)
print(inv_sort.count)