Went through the posts about the problem, but none helped me understand the issue or resolve the problem:
# This is the definition of the square() function
def square(lst1):
lst2 = []
for num in lst1:
lst2.append(num**2)
return lst2
n = [4,3,2,1]
print(list(map(square, n)))
>>>
File "test.py", line 5, in square
for num in lst1:
TypeError: 'int' object is not iterable
What is wrong with that line in the square() function definition, and what is the solution?
Thanks a lot!
map applies square to each item of your list.
So including a loop in square is redundant. lst1 is already an integer when the function gets called.
Either do:
result = square(n)
or:
result = [i*i for i in n]
the latter is better & faster than
result = list(map(square,n))
with:
def square(i):
return i*i
(or a lambda)
Related
I am new to coding in Python and I am struggling with a very simple problem. There is the same question but for javascript on the forum but it does not help me.
My code is :
def filter_list(l):
for i in l:
if i != str():
l.append(i)
i = i + 1
return(l)
print(filter_list([1,2,'a','b']))
If you can help!
thanks
Before I present solution here are some problems you need to understand.
str()
str() creates a new instance of the string class. Comparing it to an object with == will only be true if that object is the same string.
print(1 == str())
>>> False
print("some str" == str())
>>> False
print('' == str())
>>> True
iterators (no +1)
You have i = i + 1 in your loop. This doesn't make any sense. i comes from for i in l meaning i looping over the members of list l. There's no guarantee you can add 1 to it. On the next loop i will have a new value
l = [1,2,'a']
for i in l:
print(i)
>>> 1
>>> 2
>>> 'a'
To filter you need a new list
You are appending to l when you find a string. This means that when your loop finds an integer it will append it to the end of the list. And later it will find that integer on another loop interation. And append it to the end AGAIN. And find it in the next iteration.... Forever.
Try it out! See the infinite loop for yourself.
def filter_list(l):
for i in l:
print(i)
if type(i) != str:
l.append(i)
return(l)
filter_list([1,2,'a','b'])
Fix 1: Fix the type check
def filter_list(l):
for i in l:
if type(i) != str:
l.append(i)
return(l)
print(filter_list([1,2,'a','b']))
This infinite loops as discussed above
Fix 2: Create a new output array to push to
def filter_list(l):
output = []
for i in l:
if type(i) != str:
output.append(i)
return output
print(filter_list([1,2,'a','b']))
>>> [1,2]
There we go.
Fix 3: Do it in idiomatic python
Let's use a list comprehension
l = [1,2,'a','b']
output = [x for x in l if type(x) != str]
print(output)
>>> [1, 2]
A list comprehension returns the left most expression x for every element in list l provided the expression on the right (type(x) != str) is true.
I found a pyi file which has the following def
def most_common(self, n: Optional[int] = ...) -> List[Tuple[_T, int]]: ...
How could this happen? List is not defined, and no implementation?
Just highlight some valuable suggestions here for followers:
List is imported from the typing module; it's not the same thing as list. The .pyi file doesn't need to import it because stub files are never executed; they just have to be syntactically valid Python
If you use from future import annotations, you won't have to import typing to use List et al. in function annotations in .py files, either, since function annotations will be treated as string literals. (Starting in Python 4, that will be the default behavior. See PEP 563 for details.)
You are looking at the pyi file which is used solely for annotations. It is never executed by the Python interpreter. You can learn more about pyi files by reading PEP484.
Using a debugger, put a breakpoint on the line where you call most_commonand then step into the method.
Python 3.7 implementation.
...\Lib\collections\__init__.py:
def most_common(self, n=None):
'''List the n most common elements and their counts from the most
common to the least. If n is None, then list all element counts.
>>> Counter('abcdeabcdabcaba').most_common(3)
[('a', 5), ('b', 4), ('c', 3)]
'''
# Emulate Bag.sortedByCount from Smalltalk
if n is None:
return sorted(self.items(), key=_itemgetter(1), reverse=True)
return _heapq.nlargest(n, self.items(), key=_itemgetter(1))
_heapq.nlargest (in ...\Lib\heapq.py) implementation:
def nlargest(n, iterable, key=None):
"""Find the n largest elements in a dataset.
Equivalent to: sorted(iterable, key=key, reverse=True)[:n]
"""
# Short-cut for n==1 is to use max()
if n == 1:
it = iter(iterable)
sentinel = object()
if key is None:
result = max(it, default=sentinel)
else:
result = max(it, default=sentinel, key=key)
return [] if result is sentinel else [result]
# When n>=size, it's faster to use sorted()
try:
size = len(iterable)
except (TypeError, AttributeError):
pass
else:
if n >= size:
return sorted(iterable, key=key, reverse=True)[:n]
# When key is none, use simpler decoration
if key is None:
it = iter(iterable)
result = [(elem, i) for i, elem in zip(range(0, -n, -1), it)]
if not result:
return result
heapify(result)
top = result[0][0]
order = -n
_heapreplace = heapreplace
for elem in it:
if top < elem:
_heapreplace(result, (elem, order))
top, _order = result[0]
order -= 1
result.sort(reverse=True)
return [elem for (elem, order) in result]
# General case, slowest method
it = iter(iterable)
result = [(key(elem), i, elem) for i, elem in zip(range(0, -n, -1), it)]
if not result:
return result
heapify(result)
top = result[0][0]
order = -n
_heapreplace = heapreplace
for elem in it:
k = key(elem)
if top < k:
_heapreplace(result, (k, order, elem))
top, _order, _elem = result[0]
order -= 1
result.sort(reverse=True)
return [elem for (k, order, elem) in result]
[Edit: Solved, i was calling the same function from another place in the code with wrong input arguments, creating this error]
I am new to python and after some search i've decided to post my problem..
My function takes *args as input: a variable number of lists
In this function i use a for loop to read all lists and try to get len(each_list).
I then get the error TypeError: object of type 'numpy.float64' has no len()
I've tried to find the problem by myself but i don't understand the behavior
If i do:
def myfunction(* args):
for p in args:
print(isinstance(p, list))
print(type(p))
print(p)
I'll get: True + class 'list' + [value1, value2, ... etc]
But if i add one line to get the length (last one)
def myfunction(* args):
for p in args:
print(isinstance(p, list))
print(type(p))
print(p)
a = len(p)
I'll get: False + class 'numpy.float64' + error (i understand it is not iterable)
I call the function as:
All_lists = [list1, list2, list3]
myfunction(All_lists)
# I've also tried myfunction(*All_lists)
Thank you for your help
You need to do len(p) for the length of arguments list
def myfunction(* args):
for p in args:
print(isinstance(p, list))
print(type(p))
print(p)
print(len(p))
myfunction([[1,2,3],[4,5,6],[7,8,9] ])
#True
#<class 'list'>
#[1, 2, 3]
#3
I have written a code to get prime numbers upto a certain limit in a list.
As shown above.
import math
primes = []
for i in range(1, 101):
primes.append(i)
primes.remove(10) # Just removing for sake of experiment
tot = math.sqrt(len(primes))
for j in range(2, math.ceil(tot), 1):
for l in range(0, len(primes)):
k = j**2 + l*j
primes.remove(k)
primes.remove(12) # Just removing for sake of experiment
print(primes)
This code is showing error while when it removes elements from nested loop.
Error is shown above.
Traceback (most recent call last):
File "/root/PycharmProjects/love/love.py", line 13, in <module>
primes.remove(k)
ValueError: list.remove(x): x not in list
Why is this happening as this code was able to remove element which is not under nested loop but was unable to remove element which is being removed under nested loops.
Is there any alternate solution to this problem?
You are iterating over a list while you are editing a list, which is something you should never do! When you iterate the list here:
for l in range(0, len(primes)):
You are actually changing the value of len(primes) when you remove the primes! So this causes the code to act irregularly, as:
In the list comprehension, the original list is left intact, instead a new one is created. (SOURCE)
Instead, you can use list comprehension to achieve the same result!
import math
primes = []
for i in range(1, 101):
primes.append(i)
primeslst = []
def isPrime(number):
for i in range(2,int(number/2)+1):
if number%i == 0:
return True
return False
primes = [p for p in primes if not isPrime(p)]
print(primes)
Hope it helps!
I started learning Python 3.x some time ago and I wrote a very simple code which adds numbers or concatenates lists, tuples and dicts:
X = 'sth'
def adder(*vargs):
if (len(vargs) == 0):
print('No args given. Stopping...')
else:
L = list(enumerate(vargs))
for i in range(len(L) - 1):
if (type(L[i][1]) != type(L[i + 1][1])):
global X
X = 'bad'
break
if (X == 'bad'):
print('Args have different types. Stopping...')
else:
if type(L[0][1]) == int: #num
temp = 0
for i in range(len(L)):
temp += L[i][1]
print('Sum is equal to:', temp)
elif type(L[0][1]) == list: #list
A = []
for i in range(len(L)):
A += L[i][1]
print('List made is:', A)
elif type(L[0][1]) == tuple: #tuple
A = []
for i in range(len(L)):
A += list(L[i][1])
print('Tuple made is:', tuple(A))
elif type(L[0][1]) == dict: #dict
A = L[0][1]
for i in range(len(L)):
A.update(L[i][1])
print('Dict made is:', A)
adder(0, 1, 2, 3, 4, 5, 6, 7)
adder([1,2,3,4], [2,3], [5,3,2,1])
adder((1,2,3), (2,3,4), (2,))
adder(dict(a = 2, b = 433), dict(c = 22, d = 2737))
My main issue with this is the way I am getting out of the function when args have different types with the 'X' global. I thought a while about it, but I can't see easier way of doing this (I can't simply put the else under for, because the results will be printed a few times; probably I'm messing something up with the continue and break usage).
I'm sure I'm missing an easy way to do this, but I can't get it.
Thank you for any replies. If you have any advice about any other code piece here, I would be very grateful for additional help. I probably have a lot of bad non-Pythonian habits coming from earlier C++ coding.
Here are some changes I made that I think clean it up a bit and get rid of the need for the global variable.
def adder(*vargs):
if len(vargs) == 0:
return None # could raise ValueError
mytype = type(vargs[0])
if not all(type(x) == mytype for x in vargs):
raise ValueError('Args have different types.')
if mytype is int:
print('Sum is equal to:', sum(vargs))
elif mytype is list or mytype is tuple:
out = []
for item in vargs:
out += item
if mytype is list:
print('List made is:', out)
else:
print('Tuple made is:', tuple(out))
elif mytype is dict:
out = {}
for i in vargs:
out.update(i)
print('Dict made is:', out)
adder(0, 1, 2, 3, 4, 5, 6, 7)
adder([1,2,3,4], [2,3], [5,3,2,1])
adder((1,2,3), (2,3,4), (2,))
adder(dict(a = 2, b = 433), dict(c = 22, d = 2737))
I also made some other improvements that I think are a bit more 'pythonic'. For instance
for item in list:
print(item)
instead of
for i in range(len(list)):
print(list[i])
In a function like this if there are illegal arguments you would commonly short-cuircuit and just throw a ValueError.
if bad_condition:
raise ValueError('Args have different types.')
Just for contrast, here is another version that feels more pythonic to me (reasonable people might disagree with me, which is OK by me).
The principal differences are that a) type clashes are left to the operator combining the arguments, b) no assumptions are made about the types of the arguments, and c) the result is returned instead of printed. This allows combining different types in the cases where that makes sense (e.g, combine({}, zip('abcde', range(5)))).
The only assumption is that the operator used to combine the arguments is either add or a member function of the first argument's type named update.
I prefer this solution because it does minimal type checking, and uses duck-typing to allow valid but unexpected use cases.
from functools import reduce
from operator import add
def combine(*args):
if not args:
return None
out = type(args[0])()
return reduce((getattr(out, 'update', None) and (lambda d, u: [d.update(u), d][1]))
or add, args, out)
print(combine(0, 1, 2, 3, 4, 5, 6, 7))
print(combine([1,2,3,4], [2,3], [5,3,2,1]))
print(combine((1,2,3), (2,3,4), (2,)))
print(combine(dict(a = 2, b = 433), dict(c = 22, d = 2737)))
print(combine({}, zip('abcde', range(5))))