>>> import ast >>> ast.literal_eval(‘[[0,0,0], [0,0,1], [1,1,0]]’) [[0, 0, 0], [0, 0, 1], [1, 1, 0]] For tuples >>> ast.literal_eval(‘[(0,0,0), (0,0,1), (1,1,0)]’) [(0, 0, 0), (0, 0, 1), (1, 1, 0)]
Use a list comprehension: [y for x,y in A if x>2] Demo: >>> A=[(1,’A’),(2,’H’),(3,’K’),(4,’J’)] >>> [y for x,y in A if x>2] [‘K’, ‘J’] >>>
Tuples are immutable, you may not change their contents. With a list >>> x = [1,2,3] >>> x[0] = 5 >>> x [5, 2, 3] With a tuple >>> y = tuple([1,2,3]) >>> y (1, 2, 3) >>> y[0] = 5 # Not allowed! Traceback (most recent call last): File “<pyshell#20>”, line 1, in <module> … Read more
Not only is there no requirement that the objects be arranged any particular way, but many tuple implementations actually put the second object before the first one.
You can ponder this: import csv def fitem(item): item=item.strip() try: item=float(item) except ValueError: pass return item with open(‘/tmp/test.csv’, ‘r’) as csvin: reader=csv.DictReader(csvin) data={k.strip():[fitem(v)] for k,v in reader.next().items()} for line in reader: for k,v in line.items(): k=k.strip() data[k].append(fitem(v)) print data Prints: {‘Price’: [6.05, 8.05, 6.54, 6.05, 7.05, 7.45, 5.45, 6.05, 6.43, 7.05, 8.05, 3.05], ‘Type’: [‘orange’, … Read more
I would do it simply requiring a string type from each entry: from ast import literal_eval df[‘column_2’] = df.column_1.apply(lambda x: literal_eval(str(x))) If You need to advanced Exception handling, You could do, for example: def f(x): try: return literal_eval(str(x)) except Exception as e: print(e) return [] df[‘column_2’] = df.column_1.apply(lambda x: f(x))
Slight reformulation of cgeers answer to get you the tuples you want instead of arrays: var combinations = from item1 in list from item2 in list where item1 < item2 select Tuple.Create(item1, item2); (Use ToList or ToArray if you want.) In non-query-expression form (reordered somewhat): var combinations = list.SelectMany(x => list, (x, y) => Tuple.Create(x, … Read more
For a lot of data, you should consider whether you need to keep it all in a list. If you are processing each one at a time, you can create a generator that will yield each joined string, but won’t keep them all around taking up memory: new_data = (‘ ‘.join(w) for w in sixgrams) … Read more
Seems like you want itertools.combinations: >>> list(itertools.combinations((1, 2, 3), 2)) [(1, 2), (1, 3), (2, 3)] If you want sets you’ll have to convert them explicitly. If you don’t mind an iterable instead of a list, and you’re using Python 3, you can use map: >>> s = set((1, 2, 3)) >>> map(set, itertools.combinations(s, 2)) … Read more
Got it! But it need a lot of recursive magic: type PrependTuple<A, T extends Array<any>> = A extends undefined ? T : (((a: A, …b: T) => void) extends (…a: infer I) => void ? I : []) type RemoveFirstFromTuple<T extends any[]> = T[‘length’] extends 0 ? undefined : (((…b: T) => void) extends (a, … Read more