[ad_1] How about this? ArrayList<HashMap<String, HashMap<String, String>>> list = new ArrayList<>(); for (HashMap<String, HashMap<String, String>> m : list) { for (Map.Entry<String, HashMap<String, String>> e : m.entrySet()) { String key1 = e.getKey(); for (Map.Entry<String, String> me : e.getValue().entrySet()) { String key2 = me.getKey(); String value = me.getValue(); } } } Note that you really should be … Read more
[ad_1] from your question, I think (not 100% sure) you are looking for java.util.LinkedHashMap<K, V> in your case, it would be LinkedHashMap<String, Double> from java doc: Hash table and linked list implementation of the Map interface, with predictable iteration order. This implementation differs from HashMap in that it maintains a doubly-linked list running through all … Read more
[ad_1] Update: Now the situation looks a bit different as you updated the question. Here’s a quick snippet thrown together using pandas and numpy (for the sake of simplicity we replace missing ratings with zero): import numpy as np importport pandas as pd from itertools import combinations df = pd.DataFrame(critics).T.fillna(0) distances = [] for critic1, … Read more
[ad_1] You can use zip with a nested list comprehension: values = [10.5,20.2, 50.0] occ = [3,5,1] result = [x for x, y in zip(values, occ) for _ in range(y)] Alternatively, without using range: result = [elem for x, y in zip(values, occ) for elem in [x] * y] Output: [10.5, 10.5, 10.5, 20.2, 20.2, … Read more
[ad_1] a bit unclear, are you looking for this? In [41]: a = [{ ‘a’:’z’, ‘b’:’x’, ‘c’:’w’, ‘d’:’v’}, { ‘a’:’f’, ‘b’:’g’, ‘c’:’h’, ‘d’:’i’}] In [42]: a[0].get(‘a’) Out[42]: ‘z’ …or this? In [50]: a[0].values() Out[50]: [‘z’, ‘w’, ‘x’, ‘v’] …or using your provided data: In [47]: data = {‘style’: ‘-‘, ‘subCat’: ‘-‘, ‘name’: ‘Eurodollar Futures’, ‘oi’: … Read more
[ad_1] It’s somewhat unclear to me if you are intentionally trying to skip items on each pass or if that’s accidental on your part as a side effect of your code. That is, it looks like a bug to me, but maybe it’s a feature and not a bug. If I wanted to remove 5 … Read more
[ad_1] I guess you need something like this: >>> from itertools import groupby #filter items that contain ’00:00′ >>> mylist = [x for x in mylist if x[-2] != ’00:00′ ] #now group lists based on the the second last item for k,g in groupby(mylist, key = lambda x 😡 [-2]): #find the min among … Read more
[ad_1] Your answer is here: The Python Tutorial However, here you go: lists = int(raw_input(‘Introduce the number of lists you want to have: ‘)) for i in xrange(lists): numbers = int(raw_input(‘Introduce how many numbers you want it to have: ‘)) l = [] for j in xrange(numbers): l.append(int(input(‘Number: ‘))) print l 0 [ad_2] solved Python: … Read more
[ad_1] Your operator is likely wrong. Try this: bool Man::operator < (const Man & next){ if(adress < next.adress) return true; if(adress == next.adress) { if(name < next.name) return true; } return false; } 1 [ad_2] solved Sorting list. What is wrong with my sorting method? [closed]
[ad_1] You can zip the two lists and check whether both elements are > 0, then zip again to distribute the pairs to the two sub-lists >>> a= [4, 5, 4, 0, 1, 0] >>> b= [0, 4, 4, 0, 0, 0] >>> pairs = [(x, y) for (x, y) in zip(a, b) if x … Read more
[ad_1] Well designed test shows that first function is slowest on Python 2.x (mostly because two lists have to be created, first one as a increasing range, second one as a reverted first one). I also included a demo using reversed. from __future__ import print_function import sys import timeit def iterate_through_list_1(arr): lala = None for … Read more
[ad_1] You can flatten the internal dict and update it back to the original dict. Eg., product = d.pop(“product”)[0] flattened = {f”product.{key}”: value for key, value in product.items()} d.update(flattened) This should do the trick. 1 [ad_2] solved Flat a list of containing only one dict
[ad_1] The definition of valid_list should be out of the for loop, otherwise it will be overwritten. Besides, use a flag_valid to indicate whether the invalid elements exist. Try this code: from itertools import permutations def code(): valid_list = [] for line,lists in enumerate(permutations(range(4))): flag_valid = True for index,elements in enumerate(lists): if index != len(lists) … Read more
[ad_1] What you are trying to do is to unpack the elements of the list into parameters. This can be done like this: Albaqus_Function(*coord_list[0:n]) Where n is the last index+1. The *args notation is used as the following: arguments = [“arg1”, “arg1”, “arg3”] print(*arguments) This will be equivalent to: print(“arg1”, “arg2”, “arg3”) This is useful … Read more
[ad_1] Option 1 – as recommended @StefanPochmann, @rayryeng and @Clayton Wahlstrom. index = [i for (i, j) in enumerate(y) if j] print(random.sample(index, 2)) Option 2 – My original horrible implementation… import random y = [1,0,0,0,0,1,0] i = 0 index =[] for each in y: if each == 1: index.append(i) i = i + 1 print(random.sample(index, … Read more