Given a dictionary d: d # {‘A’: 1, ‘B’: 2, ‘C’: 3, ‘D’: 4} You can use a tuple to unpack the key-value pairs in the for loop header. for i, (k, v) in enumerate(d.items()): print(i, k, v) # 0 A 1 # 1 B 2 # 2 C 3 # 3 D 4 To … Read more
Use System.ComponentModel.Design.ServiceContainer that is already available in .Net framework. ServiceContainer container = new ServiceContainer(); IList<int> integers = new List<int>(); IList<string> strings = new List<string>(); IList<double> doubles = new List<double>(); container.AddService(typeof(IEnumerable<int>), integers); container.AddService(typeof(IEnumerable<string>), strings); container.AddService(typeof(IEnumerable<double>), doubles);
The values in a map may not be unique. But if they are (in your case) you can do as you wrote in your question and create a generic method to convert it: private static <V, K> Map<V, K> invert(Map<K, V> map) { Map<V, K> inv = new HashMap<V, K>(); for (Entry<K, V> entry : … Read more
As in any function call, the arguments are evaluated before the call is executed. In this case dict.get() is no exception…
If you wish to avoid rehashing the HashMap, and you know that no other elements will be placed into the HashMap, then you must take into account the load factor as well as the initial capacity. The load factor for a HashMap defaults to 0.75. The calculation to determine whether rehashing is necessary occurs whenever … Read more
I had at one time downloaded a wiktionary dump, trying to gather together words and definitions for slavic languages. I approached it using elementtree to go thru the xml file that is the dump. I would avoid trying to scrape or crawl the site, and just download the xml dump that wikimedia provides for wiktionary. … Read more
You can use the destructuring assignment. const first_dict = { ‘key_one’: ‘value_one’, ‘key_two’: ‘value_two’ } const second_dict = { ‘key_one’: ‘value_from_second_dict’, ‘key_three’: ‘value_three’ } const accumulative = { …first_dict, …second_dict } console.log(accumulative) /* [object Object] { key_one: “value_from_second_dict”, key_three: “value_three”, key_two: “value_two” } */
Use itertools.cycle to cycle around to the beginning of L2: from itertools import cycle dict(zip(L1, cycle(L2))) # {‘A’: ‘1’, ‘B’: ‘2’, ‘C’: ‘3’, ‘D’: ‘1’, ‘E’: ‘2’} In your case, concatenating L2 with itself also works. # dict(zip(L1, L2 * 2)) dict(zip(L1, L2 + L2)) # {‘A’: ‘1’, ‘B’: ‘2’, ‘C’: ‘3’, ‘D’: ‘1’, ‘E’: … Read more
When I’m working with csv files, I often use the pandas library. It makes things like this very easy. For example: import pandas as pd a = pd.read_csv(“filea.csv”) b = pd.read_csv(“fileb.csv”) b = b.dropna(axis=1) merged = a.merge(b, on=’title’) merged.to_csv(“output.csv”, index=False) Some explanation follows. First, we read in the csv files: >>> a = pd.read_csv(“filea.csv”) >>> … Read more
Dicts don’t have an order. You can call sorted but this just gives you a sorted list of the keys: >>> sorted(d) [‘a’, ‘b’, ‘c’, ‘d’] You can treat it as an iterable and sort the key-value tuples, but then you’ve just got a list of tuples. That’s not the same as a dict. >>> … Read more