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Which is more efficient in Python: list.index() or dict.get()


Which is more efficient in Python: list.index() or dict.get()

By : ASno
Date : November 22 2020, 10:56 AM
This might help you I'm trying to solve a problem on LeetCode. I'm using Python and I made an attempt as follows: ,
Is list.index() slower than dict.get()?
code :


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Most efficient way to compare words in list / dict in Python

Most efficient way to compare words in list / dict in Python


By : Justin Ashworth
Date : March 29 2020, 07:55 AM
this one helps. You can use the set data data type for all you comparisons, and the set.intersection method to get the number of matches.
It will increare algorithm efficiency, but it will only count each word once, even if it shows up in several places in the sentence.
code :
sentence = set("I love Obama and David Card, two great people. I live in a boat".split())

dico = {
'dict1':{'is','the','boat','tree'},
'dict2':{'apple','blue','red'},
'dict3':{'why','Obama','Card','two'}
}


results = {}
for key, words in dico.items():
    results[key] = len(words.intersection(sentence))
Python assignment quirk w/ list index assign, dict index assign, and dict.get

Python assignment quirk w/ list index assign, dict index assign, and dict.get


By : Viah Reyes
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Ruby and Python are different languages, and make different choices. In Python assignments are statements and evaluates multiple assignment targets from left to right. Ruby made other choices; assignments are expressions and as a result are evaluated in the opposite order.
So in Ruby this happens:
code :
if a = 42:
if a == 42:
Converting Python dictionary to List in which dict keys are at first index of the list

Converting Python dictionary to List in which dict keys are at first index of the list


By : Flor Vermassen
Date : March 29 2020, 07:55 AM
hope this fix your issue First, extract the keys from the first dictionary in your list. Then, construct lists by applying these keys to each dictionary.
code :
keys = sorted(list_of_dicts[0].keys())
table = [keys] + [[d[k] for k in keys] for d in list_of_dicts]
# [['animal', 'color'], ['lion', 'yellow'], ['elephant', 'black']]
Python: efficient way to create a list from dict values with a given order

Python: efficient way to create a list from dict values with a given order


By : sickers
Date : March 29 2020, 07:55 AM
With these it helps I have a dictionary that looks like: , Using a list comprehension is shorter:
python 3.x: which is more efficient: list of lists vs. dict?

python 3.x: which is more efficient: list of lists vs. dict?


By : Daanyaal Khan
Date : March 29 2020, 07:55 AM
I hope this helps . A 2D array
The efficient answer to storing 2D data is 2D arrays/matrixes allocated into a continuous area of memory (not like a list of lists). This avoids the multiple memory lookups required otherwise, and the calculation of a hash value at every lookup that's needed for a dict.
code :
import numpy as np
data = np.zeros( (100, 200) ) # create a 100x200 array and initialize it with zeroes
data[10,20] = 1  # set element at coordinates 10,20 to 1
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