I was trying to implement a backtracking solution to a Leetcode problem (https://leetcode.com/problems/subsets/) and found out an unexpected bug in my code. In the first solution I do out_list.append(curr_array) in line 8 and it outputs me an empty output list.
class Solution:
def subsets(self, nums):
def backtrack(curr_array, curr_idx):
if len(curr_array) > 0:
out_list.append(curr_array)
for idx in range(curr_idx, len(nums)):
curr_array.append(nums[idx])
backtrack(curr_array, idx + 1)
curr_array.pop()
out_list = []
backtrack([], 0)
return out_list
Whereas when I do out_list.append(curr_array[:]), I get the correct answer as output.
class Solution:
def subsets(self, nums):
def backtrack(curr_array, curr_idx):
if len(curr_array) > 0:
out_list.append(curr_array[:])
for idx in range(curr_idx, len(nums)):
curr_array.append(nums[idx])
backtrack(curr_array, idx + 1)
curr_array.pop()
out_list = []
backtrack([], 0)
return out_list
I've been under the impression that list1[:] is the same thing as list1. Can you tell me what am I missing here?
cur_array is a reference to the original list. When you append cur_array to out_list and then later modify cur_array, out_list changes as well.
cur_array[:] is a copy of cur_array (same as cur_array.copy()). When you append cur_array[:] to out_list and then later modify cur_array, out_list does not change because it has its own copy of cur_array.
It looks like you make a misunderstand on list.
>>> a = [] # Here we announce a list
>>> id(a)
82700416
>>> id(a[:])
82700544
As you can see, slice will make a copy of list.
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 have a tuple with tuples inside like this:
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
This goes on and on and the numbers don't matter here. The only data that matters are the names. What I need is to count how many times a given name occurs in the tuple and return a list where each item is a list and the number of times it occurs, like this:
list_that_i_want = [['Joe',2],['Kevin',1],['Donald',1]]
I don't want to use any modules or collections like Counter. I want to hard code this.
I actually wanted to hardcode the full solution and not even use the '.count()' method.
So far what I got is this:
def create_list(tuples):
new_list= list()
cont = 0
for tup in tuples:
for name in tup:
name = tup[3]
cont = tup.count(name)
if name not in new_list:
new_list.append(name)
new_list.append(cont)
return new_list
list_that_i_want = create_list(tup)
print(list_that_i_want)
And the output that I am been given is:
['Joe',1,'Kevin',1,'Donald',1]
Any help? Python newbie here.
You could. create a dictionary first and find the counts. Then convert the dictionary to a list of list.
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
dx = {}
for _,_,_,nm in tup:
if nm in dx: dx[nm] +=1
else: dx[nm] = 1
list_i_want = [[k,v] for k,v in dx.items()]
print (list_i_want)
You can replace the for_loop and the if statement section to this one line:
for _,_,_,nm in tup: dx[nm] = dx.get(nm, 0) + 1
The output will be
[['Joe', 2], ['Kevin', 1], ['Donald', 1]]
The updated code will be:
tup = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
dx = {}
for _,_,_,nm in tup: dx[nm] = dx.get(nm, 0) + 1
list_i_want = [[k,v] for k,v in dx.items()]
print (list_i_want)
Output:
[['Joe', 2], ['Kevin', 1], ['Donald', 1]]
Using an intermediary dict:
def create_list(tuple_of_tuples):
results = {}
for tup in tuple_of_tuples:
name = tup[3]
if name not in results:
results[name] = 0
results[name] += 1
return list(results.items())
Of course, using defaultdict, or even Counter, would be the more Pythonic solution.
You can try with this approach:
tuples = ((1,2,3,'Joe'),(3,4,5,'Kevin'),(6,7,8,'Joe'),(10,11,12,'Donald'))
results = {}
for tup in tuples:
if tup[-1] not in results:
results[tup[-1]] = 1
else:
results[tup[-1]] += 1
new_list = [[key,val] for key,val in results.items()]
Here, a no-counter solution:
results = {}
for t in tup:
results[t[-1]] = results[t[-1]]+1 if (t[-1] in results) else 1
results.items()
#dict_items([('Joe', 2), ('Kevin', 1), ('Donald', 1)])
I have a basic reduce function and I want to reduce a list in order to check if an item is in the list. I have defined the function below where f is a comparison function, id_ is the item I am searching for, and a is the list. For example, reduce(f, 2, [1, 6, 2, 7]) would return True since 2 is in the list.
def reduce(f, id_, a):
if len(a) == 0:
return id_
elif len(a) == 1:
return a[0]
else:
# can call these in parallel
res = f(reduce(f, id_, a[:len(a)//2]),
reduce(f, id_, a[len(a)//2:]))
return res
I tried passing it a comparison function:
def isequal(x, element):
if x == True: # if element has already been found in list -> True
return True
if x == element: # if key is equal to element -> True
return True
else: # o.w. -> False
return False
I realize this does not work because x is not the key I am searching for. I get how reduce works with summing and products, but I am failing to see how this function would even know what the key is to check if the next element matches.
I apologize, I am a bit new to this. Thanks in advance for any insight, I greatly appreciate it!
Based on your example, the problem you seem to be trying to solve is determining whether a value is or is not in a list. In that case reduce is probably not the best way to go about that. To check if a particular value is in a list or not, Python has a much simpler way of doing that:
my_list = [1, 6, 2, 7]
print(2 in my_list)
print(55 in my_list)
True
False
Edit: Given OP's comment that they were required to use reduce to solve the problem, the code below will work, but I'm not proud of it. ;^) To see how reduce is intended to be used, here is a good source of information.
Example:
from functools import reduce
def test_match(match_params, candidate):
pattern, found_match = match_params
if not found_match and pattern == candidate:
match_params = (pattern, True)
return match_params
num_list = [1,2,3,4,5]
_, found_match = reduce(test_match, num_list, (2, False))
print(found_match)
_, found_match = reduce(test_match, num_list, (55, False))
print(found_match)
Output:
True
False
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'm working out how to remove a specific item from a list.
"peppers", "cheese", "mushrooms", "bananas", "peppers"
I can locate the item "peppers", and change it to "gone!", but I really want to deleting the item, using
del blist[idx]
But that causes an error and I don't know why.
myList = ["peppers", "cheese", "mushrooms", "bananas", "peppers"]
def findInList (needle, haystack):
needle = needle.lower()
findy = []
# loops over list elements
for i in range(0, len(haystack)):
temp = haystack[i].lower()
idx = temp.find(needle)
if (idx != -1): findy.append(i)
return findy
def deleteItemInList(alist, blist):
for i in range(0, len(alist)):
idx = alist[i]
blist[idx] = "Gone!"
# del blist[idx]
# find items in list
mySearch = findInList("Peppers", myList)
# remove item from list
deleteItemInList(mySearch, myList)
print myList
Traceback: as follows
Traceback (most recent call last):
File "delete_in_list.py", line 23, in <module>
deleteItemInList(mySearch, myList)
File "delete_in_list.py", line 16, in deleteItemInList
blist[idx] = "Gone!"
IndexError: list assignment index out of range
Could someone look over the code above and point out where I'm going wrong.
You can use a list comprehension for this.
def removeWord(needle, haystack):
return [word for word in haystack if word.lower() != needle.lower()]
To find an element use this function. Or alternatively just define it as usual:
>>> find = lambda _list, query: [item.lower() for item in _list].index(query.lower())
>>> l = ['red', 'pepper']
>>> q = 'Pepper'
>>> find(l, q)
1
To remove by index just use del:
>>> del l[find(l, q)]
>>> l
['red']
I finally figured it out! Whilst iterating over the list deleting items in the list I was effectively sawing off the branch I was sitting on.
You need to loop over the list in reverse:
def deleteItemInList(alist, blist):
for i in range(len(alist) -1, -1, -1):
idx = alist[i]
del blist[idx]