#include <stdio.h> #include <stdlib.h> struct node { int data; struct node *nextPtr; }; struct node *firstPtr = NULL; void insertioon (int d){ struct node *np, *temp, *prev = NULL; int found; np=malloc(sizeof(struct node)); np->data = d; np->nextPtr = NULL; temp=firstPtr; found=0; while ((temp != NULL) && !found) { if (temp->data <d) { prev = temp; … Read more
Suppose we have 2 list x = [‘3’, ‘1’] y = [‘1’, ‘3’] Solution-1: You can simply check whether the multisets with the elements of x and y are equal: import collections collections.Counter(x) == collections.Counter(y) This requires the elements to be hashable; runtime will be in O(n), where n is the size of the lists. … Read more
itertools.izip_longest takes an optional fillvalue argument that provides the value that is used after the shorter list has been exhausted. fillvalue defaults to None, giving the behaviour you show in your question, but you can specify a different value to get the behaviour you want: fill = a[-1] if (len(a) < len(b)) else b[-1] for … Read more
def get_smallest_length(x): return [k for k in x.keys() if len(x.get(k))==min([len(n) for n in x.values()])] def get_largest_sum(x): return [k for k in x.keys() if sum(x.get(k))==max([sum(n) for n in x.values()])] x = {‘a’: [4, 2], ‘c’: [4, 3], ‘b’: [3, 4], ‘e’: [4], ‘d’: [4, 3], ‘g’: [4], ‘f’: [4]} print get_smallest_length(x) print get_largest_sum(x) Returns: [‘e’, ‘g’, … Read more
Try to do this library(‘stringr’) apply(data, 1, function(x) str_c(x$columnnane,collapse=”,”)) where, data is you dataframe and columnname is the column containing list. edited answer out = do.call(rbind, lapply(data, function(x) str_c(x,collapse=”, “))) where data is your list object if the list is stored inside a dataframe then pass the column in place of the data above like … Read more
Well, for this example, you could use list comprehension with slicing. a = [[1, 2], [1, 2, 3], [1, 2, 3, 4]] b = [x[1:] for x in a] 0 solved Remove the first items in a list of lists – make a copy instead of modifying in-place
l=[0,0,0,0,0,0,0,0,0,0,0,0] n = 11 for i in range (0,n): l[i] = 1 print l carryidx = 0 for i in reversed(range(0,n – 1)): v = l[ n – 1] + l[i] if v > 9: while l[carryidx] > 9: carryidx += 1 l[carryidx] += 1 else: l[n – 1] = v l[i] = l[i] – … Read more
Here’s something you can try. It uses a collections.defaultdict or collections.Counter to count High, Med and Low for each product, then merges the results at the end. from collections import defaultdict, Counter product_issues = [ {“product”: “battery”, “High”: 0, “Med”: 1, “Low”: 0}, {“product”: “battery”, “High”: 1, “Med”: 0, “Low”: 0}, {“product”: “battery”, “High”: 1, … Read more
rows = [(‘x’,’y’)] That mean you have a list with one element and a tuple as its element. Your list contains one element: (‘x’,’y’) at index 0. Your tuple contains two elements: ‘x’ at index 0 and ‘y’ at index 1. To have Y: rows[0][1]: rows [0] [1] ^ ^ | | List Index Tuple … Read more
Here are hundreds of ways to do this. I think one of the simplest ones is using list comprehension, like: ones = [1 for i in range(50)] 3 solved What are the possible ways to create a list containing 50 ones in python? [duplicate]
Sure, there are many ways. Simplest: def func(alist): return zip(alist, alist[1:]) This spends a lot of memory in Python 2, since zip makes an actual list and so does the slicing. There are several alternatives focused on generators that offer memory savings, such as a very simple: def func(alist): it = iter(alist) old = next(it, … Read more
Your list is separated into values. # movies: values 0. “The Holy Grail” 1. 1975 2. “Terry Jones and Terry Gilliam” 3. 91 4. [“Graham Champman”, [“Michael Palin”, “John Cleese”, “Terry Gilliam”, “Eric Idle”, “Terry Jones”]] /!\ The index begin from 0 The last value is also separated into values: # movies[4]: values 0. “Graham … Read more
Here’s a quick function to do this looping through the tuples and comparing each tuple to the label of the previous tuple and concatenating them if the labels match. def parse_tuples(x): prev_tuple = list(x[0]) parsed = [] for i in x[1:]: if i[1] == prev_tuple[1]: prev_tuple[0] += i[0] else: parsed.append(tuple(prev_tuple)) prev_tuple = list(i) parsed.append(tuple(prev_tuple)) return … Read more
Introduction This question is a duplicate of a previously asked question. This post provides an introduction to the Python programming language and how to use it to get pairs of a list. Python is a powerful and versatile programming language that can be used to solve a variety of problems. It is easy to learn … Read more
You repeatedly append the same Mutation, and end up with multiple references to it in the list. If you want different Mutations, you have to make new ones. (I assume that’s what you think is the “problem”, as you never explicitly say what is wrong about the output.) 2 solved Append not working like expected … Read more