the question goes like this: revenue = [30000, 50000, 70000, 90000] cost = [10000, 15000, 20000, 30000] Write a comprehension together with the zip() function to make a new list that maintains profits. I tried doing it with for-loop: new_list = [] for revenue, cost in zip(revenue, cost): profit = revenue – cost new_list.append(profit) new_list… Read More TypeError: 'int' object is not iterable with zip()

d = {‘Col A’: [[‘Singapore’,’Germany’,’UK’],[‘Ireland’,’Japan’,’Australia’],[‘India’,’Korea’,’Vietnam’]], ‘Col B’: [[‘Singapore’,’Germany’,’UK’],[‘Ireland’,’Japan’],[‘India’,’Mexico’,’Argentina’]]} df = pd.DataFrame(data=d) I’m trying to compare these two columns and return a new column, Col C, that contains any strings that are present in Col B but NOT present in Col A. So row 1 being the same returns no value, row 2 where A contains… Read More Compare two DataFrame columns of lists of strings (A & B) to find if any in B are NOT in A for adding to Col C

Is there a way to unpack a list of lists, but into multiple variables? scores = [[‘S’, ‘R’], [‘A’, ‘B’], [‘X’, ‘Y’], [‘P’, ‘Q’]] Into: a = [‘S’, ‘R’], b = [‘A’, ‘B’], c = [‘X’, ‘Y’], d = [‘P’, ‘Q’] Seems like something that should be quite simple and yet I’m unable to find… Read More Unpacking a list into multiple variables

I just solved a question in LeetCode, however I found that the two different list comprehensions produce different results. So what’s the difference? ans = [[0] * (n-2)] * (n-2) ans = [[0] * (n-2) for _ in range(n-2)] The question is LeetCode 2373, and the two different results are as follows: This solution gives… Read More What's the difference between the two list comprehension in Python?

list1 = [‘a’, ‘b’, ‘c’, ‘d’] list2 = [1, 2, 3, 4] # How can I get this result using list Comprehension list3 = [‘a’, ‘b’, ‘c’, ‘d’, 1, 2, 3, 4] list3 = [list1.append(x) for x in list2] print(list3) # returned [None, None, None, None] >Solution : [*list1,*list2] #[‘a’, ‘b’, ‘c’, ‘d’, 1, 2,… Read More Join two lists using list comprehension

Given the following list of tokens: a = [‘heyyo’, ‘how’, ‘ale’, ‘yiou’] And a list of tuples: b = [(‘yiou’, ‘you’), (‘heyyo’, ‘hello’)] How can I replace the elements of the list a considering the elements of the list b? For example, the expected output would be: [‘hello’, ‘how’, ‘ale’, ‘you’] This is due to… Read More How to substitute a list of words based on a list of tuples by preserving the same original order?

I would like to change values in a column for specific rows. The rows are defined in a vector containing the row indexes. I thought this could be done with list comprehension (similar to apply function in R). I tried this: [(dataframe.loc[(dataframe[‘id’]== x), ‘column_to_be_changed’] = 1) (x) for x in indexes] The error message is… Read More Set specific rows values to column in pandas dataframe using list comprehension

according to @KellyBundy answer, I’d like to get rid of repeated self-values (2, 2), (0, 0) and keep only either (2, 1) or (1, 2), keep only either (1, 0) or (0, 1) and get rid of repeated (0, 2) [(0, 2), (2, 2), (2, 1), (1, 2), (2, 2), (0,2), (2, 2), (2, 1),… Read More get rid of repeated self values and keep only one out of (2,0) (0,2)?

I am trying to create a new DataFrame column that contains words that match between a list of keywords and strings in a df column… data = { ‘Sandwich Opinions’:[‘Roast beef is overrated’,’Toasted bread is always best’,’Hot sandwiches are better than cold’] } df = pd.DataFrame(data) keywords = [‘bread’, ‘bologna’, ‘toast’, ‘sandwich’] df[‘Matches’] = [df.apply(lambda… Read More Search DataFrame column for words in list

a = [1,2,3,4,5] b = [[3,4],[4,5],[6,7]] I have two lists above. I want to compare elements of each list from list b with elements of list a, a new list is to be formed which will be a list of lists replacing the unmatched elements with ‘X. So the output should be a new list… Read More Replacing elements from list of lists when it matches elements from other list