[ad_1] items() contains tuples, key-value pairs: >>> spam.items() dict_items([(‘name’, ‘Zophie’), (‘age’, 7)]) Your key is not such a tuple. It may be contained in one of the tuples, but in does not test for containment recursively. Either test for the correct key-value tuple: >>> (‘name’, ‘Zophie’) in spam.items() True or if you can’t get access … Read more
[ad_1] First you should do is to properly define type of the graph, because unlike Python you have to specify types in Swift in declaration: var graph: [String: [String: Int]] // Dictionary(hash table) with keys of type String and values of type Dictionary<String, Int> Then you should initialize graph with some initial value, because in … Read more
[ad_1] def main(): # I’m assuming you can get this far… lines = [ ‘1,some stuff 1’, ‘2,some stuff 2,more stuff’, ‘2,some stuff 4,candy,bacon’, ‘3,some stuff 3,this,is,horrible…’ ] # Something to hold your parsed data data = {} # Iterate over each line of your file for line in lines: # Split the data apart … Read more
[ad_1] What you want is the SOWPODS word list. It has all the two-letter and three-letter words used in Scrabble. At the bottom of the Wikipedia article, you’ll find a link to a page where you can download the list. To use the word list efficiently, store it in a trie. With the help of … Read more
[ad_1] Several ways to do this, this is one: EDIT: my first solution gave a list for every value, but you only require a list when there is more than one value for a key. my_list = [‘key1=value1’, ‘key2=value2’, ‘key3=value3-1’, ‘value3-2’, ‘value3-3’, ‘key4=value4’, ‘key5=value5’, ‘value5-1’, ‘value5-2’, ‘key6=value6’] my_dict = {} current_key = None for item … Read more
[ad_1] The part I am getting hung up on is how I can loop through the values for each key and get to the individual data in each value. I can do the rest after that So if I understand you correctly then you want to iterate through a dictionary which has lists as its … Read more
[ad_1] If this is what you’re aiming for: List(List(r1), List(r2), List(r3 chain, r4), List(r5 chain, r6 chain, r7)) then here is a possibility: val rules = List(“r1”, “r2”, “r3 chain”, “r4”, “r5 chain”, “r6 chain”, “r7”) val (groups, last) = rules.foldLeft(List[List[String]](), List[String]()) { case ((groups, curGroup), rule) if rule.contains(“chain”) => (groups, rule :: curGroup) case … Read more
[ad_1] Your code works fine for me using your (slightly edited) sample data: data = [{‘id’: 1, ‘name’: ‘test’}, {‘id’: 2, ‘name’: ‘test’}, {‘id’: 3, ‘name’: ‘test’}] val = [x[‘id’] for x in data if x[‘name’] == ‘test’][0] >>> print(val) 1 However, if there is no dictionary containing a name that matches the target string: … Read more
[ad_1] Your code should be var test = [[Int: Any]]() //assign your header for tes in test{ for (key, value) in tes{ // error line print(“\(key)”) } } So first of all: You must have an array of dictionary and not just a dictionary: [[Int: Any]] Change your last for to for (key, value) in … Read more
[ad_1] PYTHON: Why does “varaible=float(variable)/len(variable2)” work, but “variable=float(variable) variable/len(variable2)” does not work? [ad_2] solved PYTHON: Why does “varaible=float(variable)/len(variable2)” work, but “variable=float(variable) variable/len(variable2)” does not work?
[ad_1] Note: Python 3.6 introduces a new, order-preserving implementation of dict, which makes the following obsolete from 3.6 onwards. Here are three iterations of your example in three different Python 3.4 interpreter sessions: Python 3.4.1 (default, Aug 8 2014, 15:05:42) [GCC 4.8.2] on linux2 Type “help”, “copyright”, “credits” or “license” for more information. >>> d={} … Read more
[ad_1] The problem is that you need to provide a lambda expression for the second argument of ToDictionary. ToDictionary also returns a Dictionary<T, U> so you won’t be able to assign it to an instance of ConcurrentDictionary<T, U>. This should do the trick: var dicFailedProxies = File.ReadLines(“failed_proxies.txt”) .Distinct() .ToDictionary(line => line, line => 0); Of … Read more
[ad_1] Update: for multiple dictionaries Iterate over your dictionaries and for every one of them, check if the value of ‘mesic’ is in [6,7,8] and if so, get the corresponding dictionary values: d1 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 8, ‘teplotaC’: 26.88} d2 = {‘stat’: u’AS’, ‘vyska’: 3.72, ‘stanice’: u’AQW00061705′, ‘mesic’: 1, ‘teplotaC’: … Read more
[ad_1] int(x) yields your error, since you can not convert “A” into an integer Correct would be: a=”ABCD” b=map(lambda x:x,a) print(list(b)) As mentioned in the comments, the following gives the same result: print(list(a)) You should probably check out some more lambda tutorials first: http://www.secnetix.de/olli/Python/lambda_functions.hawk [ad_2] solved Using lambda function in python
[ad_1] Try to loop until your check evaluates True: import time interval = 0.2 # nr of seconds while True: stop_looping = myAlertCheck() if stop_looping: break time.sleep(interval) The sleep gives you CPU time for other tasks. EDIT Ok, I’m not sure what exactly your question is. First I thought you wanted to know how to … Read more