tuples

问题 Given the target ('b', 'a') and the inputs: x0 = ('b', 'a', 'z', 'z') x1 = ('b', 'a', 'z', 'z') x2 = ('z', 'z', 'a', 'a') x3 = ('z', 'b', 'a', 'a') The aim is to find the location of the continuous ('b', 'a') element and get the output: >>> find_ba(x0) 0 >>> find_ba(x1) 0 >>> find_ba(x2) None >>> find_ba(x3) 1 Using the pairwise recipe: from itertools import tee def pairwise(iterable): "s -> (s0,s1), (s1,s2), (s2, s3), ..." a, b = tee(iterable) next(b, None) return zip(a, b) I could do this

问题 I would like to create the equivalent of: def toTupleN[A1, ..., AN, L <: HList](l: L): TupleN[A1, ..., AN] Code using toTupleN should compile only when there is exactly one N combination of values in l that the tuple could be created from. Anything else should generate a compile-time error. Available implicit conversions should be taken into account. Note that there are no restrictions on the size of l or the ordering of values in it. Example: val l = 23 :: (1, "wibble") :: (2, "wobble") ::

问题 How do I convert this dataframe location value 0 (Richmond, Virginia, nan, USA) 100 1 (New York City, New York, nan, USA) 200 to this: city state region country value 0 Richmond Virginia nan USA 100 1 New York City New York nan USA 200 Note that the location column in the first dataframe contains tuples. I want to create four columns out of the location column. 回答1: new_col_list = ['city','state','regions','country'] for n,col in enumerate(new_col_list): df[col] = df['location'].apply(lambda

问题 From this post I learned that you can concatenate tuples with: >>> tuples = (('hello',), ('these', 'are'), ('my', 'tuples!')) >>> sum(tuples, ()) ('hello', 'these', 'are', 'my', 'tuples!') Which looks pretty nice. But why does this work? And, is this optimum, or is there something from itertools that would be preferable to this construct? 回答1: the addition operator concatenates tuples in python: ('a', 'b')+('c', 'd') Out[34]: ('a', 'b', 'c', 'd') From the docstring of sum : Return the sum of

问题 I am searching some optimized datatypes for "observations-variables" table in Matlab, that can be fast and easily accessed by columns (through variables) and by rows (through observations). Here is сomparison of existing Matlab datatypes: Matrix is very fast, hovewer, it has no built-in indexing labels/enumerations for its dimensions, and you can't always remember variable name by column index. Table has very bad performance, especially when reading individual rows/columns in a for loop (I

问题 Given two lists, [a, b] and [c, d] , I'd like to get the following result: [(a,c), (a,d), (b,c), (b,d)] How can I do this in Haskell? Is there a built-in function for this, or should I implement one myself? 回答1: [ (x,y) | x<-[a,b], y<-[c,d] ] This doesn't really require any further explanation, does it? 回答2: Applicative style all the way! λ> :m + Control.Applicative λ> (,) <$> ['a','b'] <*> ['c','d'] [('a','c'),('a','d'),('b','c'),('b','d')] (I've eschewed any String syntactic sugar above, in

问题 Consider the following expressions: new Tuple<int,int>(1,2); Tuple.Create(1,2); Is there any difference between these two methods of Tuple creation? From my reading it seems to be more a convenient shorthand than anything like object creation in C++ (heap vs stack). 回答1: Well, this questions is old... but nevertheless I think I may contribute constructively. From the accepted answer: I suppose one benefit is that, since you don't have to specify the type with Tuple.Create, you can store

问题 I have a generator of tuples and I need to delete tuples containing same elements. I need this output for iterating. Input = ((1, 1), (1, 2), (1, 3), (3, 1), (3, 2), (3, 3)) Output= ((1, 1), (1, 2), (1, 3)) Order of output doesn't matter. I have checked this question but it is about lists: Delete duplicate tuples with same elements in nested list Python I use generators to achieve fastest results as the data is very large. 回答1: You can normalize the data by sorting it, then add it to a set to

问题 I have a generator of tuples and I need to delete tuples containing same elements. I need this output for iterating. Input = ((1, 1), (1, 2), (1, 3), (3, 1), (3, 2), (3, 3)) Output= ((1, 1), (1, 2), (1, 3)) Order of output doesn't matter. I have checked this question but it is about lists: Delete duplicate tuples with same elements in nested list Python I use generators to achieve fastest results as the data is very large. 回答1: You can normalize the data by sorting it, then add it to a set to

问题 very simple question: I want to do something like this: var arr1: Array[Double] = ... var arr2: Array[Double] = ... var arr3: Array[(Double,Double)] = arr1.zip(arr2) arr3.foreach(x => {if (x._1 > treshold) {x._2 = x._2 * factor}}) I tried a lot differnt syntax versions, but I failed with all of them. How could I solve this? It can not be very difficult ... Thanks! 回答1: Multiple approaches to solve this, consider for instance the use of collect which delivers an immutable collection arr4 , as