I have a 2d array, how can I delete certain element(s) from it?
x = [[2,3,4,5,2],[5,3,6,7,9,2],[34,5,7],[2,46,7,4,36]]
for i in range(len(x)):
for j in range(len(x[i])):
if x[i][j] == 2:
del x[i][j]
This will destroy the array and returns error "list index out of range".
you can use pop on the list item. For example -
>>> array = [[1,2,3,4], [6,7,8,9]]
>>> array [1].pop(3)
>>> array
[[1, 2, 3, 4], [6, 7, 8]]
I think this can solve your problem.
x = [[2,3,4,5,2],[5,3,6,7,9,2],[34,5,7],[2,46,7,4,36]]
for i in range(len(x)):
for j in range(len(x[i])):
if j<len(x[i]):
if x[i][j] == 2:
del x[i][j]
I have tested it locally and working as expected.Hope it will help.
Mutating a list while iterating over it is always a bad idea. Just make a new list and add everything except those items you want to exclude. Such as:
x = [[2,3,4,5,2],[5,3,6,7,9,2],[34,5,7],[2,46,7,4,36]]
new_array = []
temp = []
delete_val = 2
for list_ in x:
for element in list_:
if element != delete_val:
temp.append(element)
new_array.append(temp)
temp = []
x = new_array
print(x)
Edit: made it a little more pythonic by omitting list indices.
I think this is more readable at the cost of temporarily more memory usage (making a new list) compared to the solution that Sai prateek has offered.
Related
a = [0,0,0,0,0,0,0,0,0,1,2,3,4,0,0]
for i in a:
if i < 3:
del a[0]
print(a)
[0, 0, 0, 1, 2, 3, 4]
should be:
[3,4,0,0]
You are using i in a confusing way, since you're not using it as an index (which is kinda the norm) but as an iterator. Just for readability I would suggest changing the iterator name to something more descriptive.
If I understand you're question correctly, you want to delete the first element, until it is bigger or equal 3. If that is your question, you could do it like this:
a = [0,0,0,0,0,0,0,0,0,1,2,3,4,0,0]
while a[0]<3:
del a[0]
print(a) # [3,4,0,0]
a = [0,0,0,0,0,0,0,0,0,1,2,3,4]
n= []
for i in range(len(a)):
if(a[i]<3):
continue
else:
n.append(a[i])
print(n) # [3,4]
I am having list which as follows
input_list= [2, 3, 5, 2, 5, 1, 5]
I want to get all the indexes of maximum value. Need efficient solution. The output will be as follows.
output = [2,4,6] (The above list 5 is maximum value in a list)
I have tried by using below code
m = max(input_list)
output = [i for i, j in enumerate(a) if j == m]
I need to find any other optimum solution.
from collections import defaultdict
dic=defaultdict(list)
input_list=[]
for i in range(len(input_list)):
dic[input_list[i]]+=[i]
max_value = max(input_list)
Sol = dic[max_value]
You can use numpy (numpy arrays are very fast):
import numpy as np
input_list= np.array([2, 3, 5, 2, 5, 1, 5])
i, = np.where(input_list == np.max(input_list))
print(i)
Output:
[2 4 6]
Here's the approach which is described in comments. Even if you use some library, fundamentally you need to traverse at least once to solve this problem (considering input list is unsorted). So even lower bound for the algorithm would be Omega(size_of_list). If list is sorted we can leverage binary_search to solve the problem.
def max_indexes(l):
try:
assert l != []
max_element = l[0]
indexes = [0]
for index, element in enumerate(l[1:]):
if element > max_element:
max_element = element
indexes = [index + 1]
elif element == max_element:
indexes.append(index + 1)
return indexes
except AssertionError:
print ('input_list in empty')
Use a for loop for O(n) and iterating just once over the list resolution:
from itertools import islice
input_list= [2, 3, 5, 2, 5, 1, 5]
def max_indexes(l):
max_item = input_list[0]
indexes = [0]
for i, item in enumerate(islice(l, 1, None), 1):
if item < max_item:
continue
elif item > max_item:
max_item = item
indexes = [i]
elif item == max_item:
indexes.append(i)
return indexes
Here you have the live example
Think of it in this way, unless you iterate through the whole list once, which is O(n), n being the length of the list, you won't be able to compare the maximum with all values in the list, so the best you can do is O(n), which you already seems to be doing in your example.
So I am not sure you can do it faster than O(n) with the list approach.
Lets say I have a list with three arrays as following:
[(1,2,0),(2,9,6),(2,3,6)]
Is it possible I get the average by diving each "slot" of the arrays in the list.
For example:
(1+2+2)/3, (2+0+9)/3, (0+6+6)/3
and make it become new arraylist with only 3 integers.
You can use zip to associate all of the elements in each of the interior tuples by index
tups = [(1,2,0),(2,9,6),(2,3,6)]
print([sum(x)/len(x) for x in zip(*tups)])
# [1.6666666666666667, 4.666666666666667, 4.0]
You can also do something like sum(x)//len(x) or round(sum(x)/len(x)) inside the list comprehension to get an integer.
Here are couple of ways you can do it.
data = [(1,2,0),(2,9,6),(2,3,6)]
avg_array = []
for tu in data:
avg_array.append(sum(tu)/len(tu))
print(avg_array)
using list comprehension
data = [(1,2,0),(2,9,6),(2,3,6)]
comp = [ sum(i)/len(i) for i in data]
print(comp)
Can be achieved by doing something like this.
Create an empty array. Loop through your current array and use the sum and len functions to calculate averages. Then append the average to your new array.
array = [(1,2,0),(2,9,6),(2,3,6)]
arraynew = []
for i in range(0,len(array)):
arraynew.append(sum(array[i]) / len(array[i]))
print arraynew
As you were told in the comments with sum and len it's pretty easy.
But in python I would do something like this, assuming you want to maintain decimal precision:
list = [(1, 2, 0), (2, 9, 6), (2, 3, 6)]
res = map(lambda l: round(float(sum(l)) / len(l), 2), list)
Output:
[1.0, 5.67, 3.67]
But as you said you wanted 3 ints in your question, would be like this:
res = map(lambda l: sum(l) / len(l), list)
Output:
[1, 5, 3]
Edit:
To sum the same index of each tuple, the most elegant method is the solution provided by #PatrickHaugh.
On the other hand, if you are not fond of list comprehensions and some built in functions as zip is, here's a little longer and less elegant version using a for loop:
arr = []
for i in range(0, len(list)):
arr.append(sum(l[i] for l in list) / len(list))
print(arr)
Output:
[1, 4, 4]
I need to print only duplicate numbers in a list and need to multiply by count. the code is as follows , the output should be ,
{1:3, 2:2, 3:2} need to multiply each numbers by count and print as separate answers:
answer1 = 1*3, answer2 = 2*2 , answer3 = 3*2
Current attempt:
from collections import Counter
alist = [1,2,3,5,1,2,1,3,1,2]
a = dict(Counter(a_list))
print(a)
Counter already does the heavy lifting. So for the rest, what about generating a list of the values occuring more than once, formatting the output as you wish ? (sorting the keys seems necessary so indexes match the keys order):
from collections import Counter
a_list = [1,2,3,5,1,2,1,3,1,2]
a = ["{}*{}".format(k,v) for k,v in sorted(Counter(a_list).items()) if v > 1]
print(a)
result:
['1*4', '2*3', '3*2']
If you want the numerical result instead:
a = [k*v for k,v in sorted(Counter(a_list).items()) if v > 1]
result (probably more useful):
[4, 6, 6]
Assigning to separate variables (answer1,answer2,answer3 = a) is not a very good idea. Keep a indexed list
In a list, there might be several largest numbers. I want to get the indices of them all.
For example:
In the list a=[1,2,3,4,5,5,5]
The indices of the largest numbers are 4,5,6
I know the question is easy for most of people, but please be patient to answer my question.
Thanks :-)
In [156]: a=[1,2,3,4,5,5,5]
In [157]: m = max(a)
In [158]: [i for i,num in enumerate(a) if num==m]
Out[158]: [4, 5, 6]
1) create a variable maxNum = 0
2) loop through list if a[i] > maxNum : maxNum = a[i]
3)loop through list a second time now if a[i] == maxNum: print(i)
Try this:
a=[1,2,3,4,5,5,5]
b = max(a)
[x for x, y in enumerate(a) if y == b]
Use heapq:
import heapq
from operator import itemgetter
a=[1,2,3,4,5,5,5]
largest = heapq.nlargest(3, enumerate(a), key=itemgetter(1))
indices, _ = zip(*largest)
Of course, if your list is already sorted (your example list is), it may be as simple as doing
indices = range(len(a) - 3, len(a))
mylist=[1,2,3,3,3]
maxVal=max(mylist)
for i in range(0,len(mylist)):
if(mylist[i]==maxVal):
print i