1 a -[ ] ab - [ ) s for i in (Data.head(200) and Data1.tail(200)): 16 if i[-2] not in a: a.append(i[-1]) b.append(i[-2]) 3 else : b[a.index(1) + -i[a] trans - max[b] ecusid = a[b.index(trans(a)) ] 1 print("Hurray. The most number of transactions is done by Customer 18:", cusida" .He has made ", trans" , transactions. He wi File "", line 15 b[a.index(1) + -i[a] ntaxError: invalid syntax
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- 185e def find(self, key): 186e 187 188 Efficiency: 189 Finds and returns a copy of value in the set that matches key. Use: value = source.find(key) 190 191 192 193 Parameters: 194 key - a partial data element (?) 195 Returns: 196 value - a copy of the full value matching key, otherwise None (?) 197 198 199 200 # your code here 201An array may contain multiple duplicate values, as shown below. In order to design an algorithm with the best average performance to remove all duplicate values, which of the following data structures should be used? 110, 1100, s, 10, s6, s, 96, 34, s, 34, 20, 110, 30 Question 16 options: hash table stack list priority queueIntSet unionWith(const IntSet& otherIntSet) const; //Unions the current set with the passed set. IntSet::unionWith(const IntSet& otherIntSet){ for (int i = 0; i < otherIntSet.used; ++i) //For each element in the passed set. { add(otherIntSet.data[i]); //Add it to current set. }}
- #The Iris Dataset import sklearn.datasetsimport matplotlib.pyplot as plt import numpy as np import scipy iris = sklearn.datasets.load_iris() Write a function that takes in an index i and prints out a verbose desciption of the species and measurements for data point i. For example:Data point 5 is of the species setosaIts sepal length (cm) is 5.4Its sepal width (cm) is 3.9Its petal length (cm) is 1.7Its petal width (cm) is 0.4def small_index(items: list[int]) -> int:"""Return the index of the first integer in items that is less than its index,or -1 if no such integer exists in items.>>> small_index([2, 5, 7, 99, 6])-1>>> small_index([-5, 8, 9, 16])0>>> small_index([5, 8, 9, 0, 1, 3])3"""def drop_uninformative_objects(X, y):# your code herereturn X_subset, y_subset # TEST drop_uninformative_objects functionA = pd.DataFrame(np.array([[0, 3, np.nan],[4, np.nan, np.nan],[np.nan, 6, 7],[np.nan, np.nan, np.nan],[5, 5, 5],[np.nan, 8, np.nan],]))b = pd.Series(np.arange(6))A_subset, b_subset = drop_uninformative_objects(A, b) assert A_subset.shape == (3, 3)assert b_subset.shape == (3,)
- void listEmployees (void) { for (int i=0; i 10000. Make a guess about why the comparison function takes 2 struct Employee parameters (as opposed to struct Employee *) **The following code doesn't correctly return the index values of the minimum contiguous subsequence. Please modify it so that it does, with comments if possible. def minSumRec(a, left, right): if left == right: return [a[left], left, right] center = int((left + right) / 2) min_left_sum = minSumRec(a, left, center) min_right_sum = minSumRec(a, center + 1, right) min_left_border_sum, left_border_sum = 1e9, 0 for i in range(center, left - 1, -1): left_border_sum += a[i] if left_border_sum < min_left_border_sum: min_left_border_sum = left_border_sum min_right_border_sum, right_border_sum = 1e9, 0 for i in range(center + 1, right + 1): right_border_sum += a[i] if right_border_sum < min_right_border_sum: min_right_border_sum = right_border_sum min_border_sum = min_right_border_sum + min_left_border_sum minSum = min(min_left_sum[0], min_right_sum[0], min_border_sum) if minSum == min_left_sum: return min_left_sum if minSum == min_right_sum: return min_right_sum return…def large_matrix(matrix: list[list[int]]) -> int:"""Returns the area of the rectangle in the area of a rectangle is defined by number of 1's that it contains.The matrix will only contain the integers 1 and 0.>>> case = [[1, 0, 1, 0, 0],... [1, 0, 1, 1, 1],... [1, 1, 1, 1, 1],... [1, 0, 0, 1, 0]]>>> largest_in_matrix(case1)6"" You must use this helper code: def large_position(matrix: list[list[int]], row: int, col: int) -> int: a = rowb = colmax = 0temp = 0rows = len(matrix)column = len(matrix[a])while a < rows and matrix[a][col] == 1:temp = 0while b < column and matrix[a][b] == 1:temp = temp + 1b = b + 1column = ba = a + 1if (a != row+1):temp2 = temp * (a - row)else:temp2 = tempif max < temp2:max = temp2b = colreturn max """ remember: Please do this on python you should not use any of the following: dictionaries or dictionary methods try-except break and continue statements recursion map / filter import""'
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