for each_word1 in list_a:
###compare all values to...###
for each_word2 in list_b:
if each_word2 in list_b == any_word in list_a:
add each_word2 to list_c
###something like that###
Python does have the figure of sets which are ideal for situation like this. These are data structures that excel in fast comparison of containment, and has the same primitive operations we have in basic Math back in the first school grades.
Just convert one of your lists to a set (this will remove duplcates if any), and use the intersection operation. Convert the result back to a list if you need:
list_c = list(set(list_a).intersection(list_b))
This is what I would do (if I understand you correctly).
list_a = ['abc', 'def', 'ghi']
list_b = ['ghi', 'jkl', 'mno']
list_c = []
for string in list_a:
if string in list_b:
list_c.append(string)
Related
Lets say I have two 2D lists like this:
list1 = [ ['A', 5], ['X', 7], ['P', 3]]
list2 = [ ['B', 9], ['C', 5], ['A', 3]]
I want to compare these two lists and find where the 2nd item matches between the two lists e.g here we can see that numbers 5 and 3 appear in both lists. The first item is actually not relevant in comparison.
How do I compare the lists and copy those values that appear in 2nd column of both lists? Using 'x in list' does not work since these are 2D lists. Do I create another copy of the lists with just the 2nd column copied across?
It is possible that this can be done using list comprehension but I am not sure about it so far.
There might be a duplicate for this but I have not found it yet.
The pursuit of one-liners is a futile exercise. They aren't always more efficient than the regular loopy way, and almost always less readable when you're writing anything more complicated than one or two nested loops. So let's get a multi-line solution first. Once we have a working solution, we can try to convert it to a one-liner.
Now the solution you shared in the comments works, but it doesn't handle duplicate elements and also is O(n^2) because it contains a nested loop. https://wiki.python.org/moin/TimeComplexity
list_common = [x[1] for x in list1 for y in list2 if x[1] == y[1]]
A few key things to remember:
A single loop O(n) is better than a nested loop O(n^2).
Membership lookup in a set O(1) is much quicker than lookup in a list O(n).
Sets also get rid of duplicates for you.
Python includes set operations like union, intersection, etc.
Let's code something using these points:
# Create a set containing all numbers from list1
set1 = set(x[1] for x in list1)
# Create a set containing all numbers from list2
set2 = set(x[1] for x in list2)
# Intersection contains numbers in both sets
intersection = set1.intersection(set2)
# If you want, convert this to a list
list_common = list(intersection)
Now, to convert this to a one-liner:
list_common = list(set(x[1] for x in list1).intersection(x[1] for x in list2))
We don't need to explicitly convert x[1] for x in list2 to a set because the set.intersection() function takes generator expressions and internally handles the conversion to a set.
This gives you the result in O(n) time, and also gets rid of duplicates in the process.
Provided with a list of lists. Here's an example myList =[[70,83,90],[19,25,30]], return a list of lists which contains the difference between the elements. An example of the result would be[[13,7],[6,5]]. The absolute value of (70-83), (83-90), (19-25), and (25-30) is what is returned. I'm not sure how to iterate through the list to subtract adjacent elements without already knowing the length of the list. So far I have just separated the list of lists into two separate lists.
list_one = myList[0]
list_two = myList[1]
Please let me know what you would recommend, thank you!
A custom generator can return two adjacent items at a time from a sequence without knowing the length:
def two(sequence):
i = iter(sequence)
a = next(i)
for b in i:
yield a,b
a = b
original = [[70,83,90],[19,25,30]]
result = [[abs(a-b) for a,b in two(sequence)]
for sequence in original]
print(result)
[[13, 7], [6, 5]]
Well, for each list, you can simply get its number of elements like this:
res = []
for my_list in list_of_lists:
res.append([])
for i in range(len(my_list) - 1):
# Do some stuff
You can then add the results you want to res[-1].
List of integer value passed through input function and then stored in a list. After which performing the operation to find the sum of all the numbers in the list
lst = list( input("Enter the list of items :") )
sum_element = 0
for i in lst:
sum_element = sum_element+int(i)
print(sum_element)
Say you want to create a list with 8 elements. By writing list(8) you do not create a list with 8 elements, instead you create the list that has the number 8 as it's only element. So you just get [8].
list() is not a Constructor (like what you might expect from other languages) but rather a 'Converter'. And list('382') will convert this string to the following list: ['3','8','2'].
So to get the input list you might want to do something like this:
my_list = []
for i in range(int(input('Length: '))):
my_list.append(int(input(f'Element {i}: ')))
and then continue with your code for summation.
A more pythonic way would be
my_list = [int(input(f'Element {i}: '))
for i in range(int(input('Length: ')))]
For adding all the elements up you could use the inbuilt sum() function:
my_list_sum = sum(my_list)
lst=map(int,input("Enter the elements with space between them: ").split())
print(sum(lst))
Can someone explain the last line of this Python code snippet to me?
Cell is just another class. I don't understand how the for loop is being used to store Cell objects into the Column object.
class Column(object):
def __init__(self, region, srcPos, pos):
self.region = region
self.cells = [Cell(self, i) for i in xrange(region.cellsPerCol)] #Please explain this line.
The line of code you are asking about is using list comprehension to create a list and assign the data collected in this list to self.cells. It is equivalent to
self.cells = []
for i in xrange(region.cellsPerCol):
self.cells.append(Cell(self, i))
Explanation:
To best explain how this works, a few simple examples might be instructive in helping you understand the code you have. If you are going to continue working with Python code, you will come across list comprehension again, and you may want to use it yourself.
Note, in the example below, both code segments are equivalent in that they create a list of values stored in list myList.
For instance:
myList = []
for i in range(10):
myList.append(i)
is equivalent to
myList = [i for i in range(10)]
List comprehensions can be more complex too, so for instance if you had some condition that determined if values should go into a list you could also express this with list comprehension.
This example only collects even numbered values in the list:
myList = []
for i in range(10):
if i%2 == 0: # could be written as "if not i%2" more tersely
myList.append(i)
and the equivalent list comprehension:
myList = [i for i in range(10) if i%2 == 0]
Two final notes:
You can have "nested" list comrehensions, but they quickly become hard to comprehend :)
List comprehension will run faster than the equivalent for-loop, and therefore is often a favorite with regular Python programmers who are concerned about efficiency.
Ok, one last example showing that you can also apply functions to the items you are iterating over in the list. This uses float() to convert a list of strings to float values:
data = ['3', '7.4', '8.2']
new_data = [float(n) for n in data]
gives:
new_data
[3.0, 7.4, 8.2]
It is the same as if you did this:
def __init__(self, region, srcPos, pos):
self.region = region
self.cells = []
for i in xrange(region.cellsPerCol):
self.cells.append(Cell(self, i))
This is called a list comprehension.
I want to match two lists from which one list is smaller while other is a bigger one. If a match occurs between two lists then put the matching element in a new list at the same index instead of putting it another index. You can understand my question from the code given below:
list1=['AF','KN','JN','NJ']
list2=['KNJ','NJK','JNJ','INS','AFG']
matchlist = []
smaller_list_len = min(len(list1),len(list2))
for ind in range(smaller_list_len):
elem2 = list1[ind]
elem1 = list2[ind][0:2]
if elem1 in list2:
matchlist.append(list1[ind])
Obtained output
>>> matchlist
['KNJ', 'NJK', 'JNJ']
Desired Output
>>> matchlist
['AFG', 'KNJ', 'JNJ', 'NJK']
Is there a way to get the desired output?
Use a nested loop iterating over the 3-char list. When an item in that list contains the current item in the 2-char list, append it and break out of the inner loop:
list1=['AF','KN','JN','NJ']
list2=['KNJ','NJK','JNJ','INS','AFG']
matchlist = []
smaller_list_len = min(len(list1),len(list2))
for ind in range(smaller_list_len):
for item in list2:
if list1[ind] in item:
matchlist.append(item)
break
Given the question doesn't specify any constraints, in a more pythonic way, using a list comprehension:
list1=['AF','KN','JN','NJ']
list2=['KNJ','NJK','JNJ','INS','AFG']
matchlist=[e2 for e1 in list1 for e2 in list2 if e2.startswith(e1)]
produces
['AFG', 'KNJ', 'JNJ', 'NJK']