dictionary

[Solved] Why do keys() and items() methods return different Boolean values for same key? (Python 3.6) [closed]

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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 to … Read more

[Solved] How to implement this Python code in Swift?

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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 Swift … Read more

[Solved] Sorting strings into dictionary, where initial char is the key and the value is a list of all lines starting with that char

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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 on … Read more

[Solved] Python convert list to dict with multiple key value [closed]

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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 in … Read more

[Solved] scala how do you group elements in a map

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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 ((groups, … Read more

[Solved] Python: return first value by condition of nested dictionary in array [closed]

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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

[Solved] PYTHON: Why does “varaible=float(variable)/len(variable2)” work, but “variable=float(variable) variable/len(variable2)” does not work?

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PYTHON: Why does “varaible=float(variable)/len(variable2)” work, but “variable=float(variable) variable/len(variable2)” does not work? solved PYTHON: Why does “varaible=float(variable)/len(variable2)” work, but “variable=float(variable) variable/len(variable2)” does not work?

[Solved] Do string representations of dictionaries have order in Python 3.4?

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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

[Solved] How to read lines to dictionary with linQ [closed]

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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 course, … Read more

[Solved] Python dictionary subset

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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’: 26.88} … Read more

[Solved] Using lambda function in python

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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 solved Using lambda function in python