Which method is faster for Python iterating?

Which method is faster for Python iterating?

By : user2195
Date : November 22 2020, 10:31 AM
Hope this helps Try to timeit both.
But second is widely known to be faster.
code :
import random
import timeit

old_list = [ random.randint(0, 100000) for i in range(0, 100) ]

def floop(old_list):
    new_list = []
    for value in old_list:
    return new_list

def lcomp(old_list):
    new_list = [ value for value in old_list ]
    return new_list

if __name__=='__main__':
    results_floop = timeit.Timer('floop(old_list)', "from __main__ import floop, old_list").timeit()
    results_lcomp = timeit.Timer('lcomp(old_list)', "from __main__ import lcomp, old_list").timeit()
    print("Function\t\tSeconds elapsed")
    print("For loop\t\t{}".format(results_floop))
    print("List comp\t\t{}".format(results_lcomp))
~/python » python3 lists.py
Function        Seconds elapsed
For loop        11.089475459069945
List comp       5.985794545034878

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Iterating over a list, modifying each element: is there a faster method?

Iterating over a list, modifying each element: is there a faster method?

By : Phone Myint Kyaw
Date : March 29 2020, 07:55 AM
wish of those help Nope, you're good. That's about as efficient as it'll get. There's no magic that avoids iterating.
One point to keep in mind, 'though: If listToTrim is not a random-access-list (i.e. it does not implement RandomAccess), then using an Iterator (or an enhanced for-loop, which uses an Iterator internally) instead of a traditional for-loop is usually much more efficient. The most notable List that doesn't implement RandomAccess is the LinkedList. Calling l.get(300) on a LinkedList with 600 elements will have to iterate through ~300 elements to get the correct one!
code :
public List<String> trimStrings(Listy<String> listToTrim) {
    List<String> trimmedList = new ArrayList<String>(listToTrim.size());
    for (String str : listToTrim) {
    return trimmedList;
public void trimStringsInPlace(List<String> listToTrim) {
    ListIterator<String> it = listToTrim.listIterator();
    while (it.hasNext()) {
Python: why is it faster to iterate over a list than iterating over an xrange() generator define on its length?

Python: why is it faster to iterate over a list than iterating over an xrange() generator define on its length?

By : JhayCo
Date : March 29 2020, 07:55 AM
I hope this helps you . I did it in Python3 , but the same results arose. I put the range creation in setup for a more accurate comparison
code :
In [1]: timeit.timeit('for _ in a: continue', setup='a=list(range(10000))', number=10000)
Out[1]: 1.195666481000444

In [2]: timeit.timeit('for _ in a: continue', setup='a=range(10000)', number=10000)
Out[2]: 2.4083170039994
In [3]: timeit.timeit('for _ in range(10000): continue', number=10000)
Out[3]: 4.166428555001403

In [4]: timeit.timeit('for _ in list(range(10000)): continue', number=10000)
Out[4]: 5.800707030000922
iterating a single item list faster than iterating a long string? #Python #Cherrypy

iterating a single item list faster than iterating a long string? #Python #Cherrypy

By : Jeffrey Bacon
Date : March 29 2020, 07:55 AM
wish helps you I think that code only makes sense if you recognize that prior to the code with that comment, self.body could be either a single string, or an iterable sequence that contains many strings. Other code will use it as the latter (iterating on it and doing string stuff with the items).
While would technically work to let that later code loop over the characters of the single string, processing the data character by character is likely inefficient. So the code below the comment wraps a list around the single string, letting it get processed all at once.
Is there a faster way to check if all objects in a python list are int instances (or some other class) than iterating ov

Is there a faster way to check if all objects in a python list are int instances (or some other class) than iterating ov

By : Mr. Pipapo
Date : March 29 2020, 07:55 AM
wish helps you You can combine all() with isinstance() like this
code :
a = [1, 2, 3, 4]

# return true
print(all(isinstance(i, int) for i in a))

b = [1, 2, 3, 4.5]

# return false
print(all(isinstance(i, int) for i in b))

#for list of instances
c = ['hello', 2, 3, 4]

#return true
print(all(isinstance(i, (int, str)) for i in c))
Faster way of iterating through lists in Python

Faster way of iterating through lists in Python

By : Kangraejae
Date : September 24 2020, 12:00 AM
will be helpful for those in need I have list vector_list of length 800,000, where the elements are lists of size 768. I'm trying to add 768 columns to a pandas dataframe where each column is 800,000 long and represents an element from each list. Here's my code: , Directly pass the list to DataFrame constructor.
code :
import pandas as pd
_list = [[1, 2], [3, 4], [5, 6], [7, 8]] 
df = pd.DataFrame(_list) 
   0  1
0  1  2
1  3  4
2  5  6
3  7  8
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