I am generating lists of arguments of this sort:
list = {{i_1,min_1,max_1},{i_2,min_2,max_2}, ... ,{i_n,min_n,max_n}}
How can I use this kind of lists as an argument for the multiple Sum?
Sum[f,{i_1,min_1,max_1},{i_2,min_2,max_2}, ... ,{i_n,min_n,max_n}]
Sum has attribute HoldAll so you must force evaluation of Sequence with Evaluate.
list = {{i1, min1, max1}, {i2, min2, max2}, {in, minn, maxn}};
Sum[f, Evaluate[Sequence ## list]]
You can't use underscore in Mathematica for variable names as it indicates a pattern.
Hope this helps.
Related
I am trying to write a code which calculates the HCF of two numbers but I am either getting a error or an empty list as my answer
I was expecting the HCF, My idea was to get the factors of the 2 given numbers and then find the common amongst them then take the max out of that
For future reference, do not attach screenshots. Instead, copy your code and put it into a code block because stack overflow supports code blocks. To start a code block, write three tildes like ``` and to end it write three more tildes to close. If you add a language name like python, or javascript after the first three tildes, syntax highlighting will be enabled. I would also create a more descriptive title that more accurately describes the problem at hand. It would look like so:
Title: How to print from 1-99 in python?
for i in range(1,100):
print(i)
To answer your question, it seems that your HCF list is empty, and the python max function expects the argument to the function to not to be empty (the 'arg' is the HCF list). From inspection of your code, this is because the two if conditions that need to be satisfied before anything is added to HCF is never satisfied.
So it could be that hcf2[x] is not in hcf and hcf[x] is not in hcf[x] 2.
What I would do is extract the logic for the finding of the factors of each number to a function, then use built in python functions to find the common elements between the lists. Like so:
num1 = int(input("Num 1:")) # inputs
num2 = int(input("Num 2:")) # inputs
numberOneFactors = []
numberTwoFactors = []
commonFactors = []
# defining a function that finds the factors and returns it as a list
def findFactors(number):
temp = []
for i in range(1, number+1):
if number%i==0:
temp.append(i)
return temp
numberOneFactors = findFactors(num1) # populating factors 1 list
numberTwoFactors = findFactors(num2) # populating factors 2 list
# to find common factors we can use the inbuilt python set functions.
commonFactors = list(set(numberOneFactors).intersection(numberTwoFactors))
# the intersection method finds the common elements in a set.
I am a beginner , trying to understand how list comprehension for multiple input works.
Can someone explain how the below code works?
x,y = [int(x) for x in input("Enter the value ").split()]
print(x,y)
Thanks in advance!
This is actually is not directly related to list comprehensions but instead a concept called "sequence unpacking", which applies to any sequence type (list, tuple, range). What is happening here is that the user input is expected to be two whitespace-separated values. The split call will split the user input on the whitespace, returning a list of size 2. Then, the list comprehension is looping over each element of this split-produced list and converting each one to an int. Thus, the list comprehension will return a list of length 2, and each of its elements will be "unpacked" separately into the x and y variables on the left-hand side of the assignment operator. Here is an excerpt from the Data Structures section of the Python tutorial that explains sequence unpacking:
The statement t = 12345, 54321, 'hello!' is an example of tuple packing: the values 12345, 54321 and 'hello!' are packed together in a tuple. The reverse operation is also possible:
>>> x, y, z = t
This is called, appropriately enough, sequence unpacking and works for
any sequence on the right-hand side. Sequence unpacking requires that
there are as many variables on the left side of the equals sign as
there are elements in the sequence. Note that multiple assignment is
really just a combination of tuple packing and sequence unpacking.
Note that this only works if the user input is of length 2, else the
sequence unpacking will not work and will result in an error.
Seems like some use to knowing a good pattern to make an n-step composition or pipeline from a binary function. Maybe it's obvious or common knowledge.
What I was trying to do was R.either(predicate1, predicate2, predicate3, ...) but R.either is one of these binary functions. I thought R.composeWith might be part of a good solution but didn't get it to work right. Then I think R.o is at the heart of it, or perhaps R.chain somehow.
Maybe there's a totally different way to make an n-ary either that could be better than a "compose-with"(R.either)... interested if so but trying to ask a more general question than that.
One common way for converting a binary function into one that takes many arguments is by using R.reduce. This requires at least the arguments of the binary function and its return type to be the same type.
For your example with R.either, it would look like:
const eithers = R.reduce(R.either, R.F)
const fooOr42 = eithers([ R.equals("foo"), R.equals(42) ])
This accepts a list of predicate functions that will each be given as arguments to R.either.
The fooOr42 example above is equivalent to:
const fooOr42 = R.either(R.either(R.F, R.equals("foo")), R.equals(42))
You can also make use of R.unapply if you want to convert the function from accepting a list of arguments, to a variable number of arguments.
const eithers = R.unapply(R.reduce(R.either, R.F))
const fooOr42 = eithers(R.equals("foo"), R.equals(42))
The approach above can be used for any type that can be combined to produce a value of the same type, where the type has some "monoid" instance. This just means that we have a binary function that combines the two types together and some "empty" value, which satisfy some simple laws:
Associativity: combine(a, combine(b, c)) == combine(combine(a, b), c)
Left identity: combine(empty, a) == a
Right identity: combine(a, empty) == a
Some examples of common types with a monoid instance include:
arrays, where the empty list is the empty value and concat is the binary function.
numbers, where 1 is the empty value and multiply is the binary function
numbers, where 0 is the empty value and add is the binary function
In the case of your example, we have predicates (a function returning a boolean value), where the empty value is R.F (a.k.a (_) => false) and the binary function is R.either. You can also combine predicates using R.both with an empty value of R.T (a.k.a (_) => true), which will ensure the resulting predicate satisfies all of the combined predicates.
It is probably also worth mentioning that you could alternatively just use R.anyPass :)
I'm trying to explore kotlin but when i came across list of strings like below.
I tried min and max functions with it.
And i initially thought it will give compile time error but i didn't get that.
And when i print min i got a555585887996669 as output which is the longest word in array.
val list = listOf<String>("a555585887996669","abtfcr6cr","abcde","abcd")
println(list.min()) //a555585887996669
I need to know on what basis it is returning this value
why min and max is supported to list of strings
The min() and max() extension functions operate on anything that can be compared. That includes numeric types, but also on anything that implements the Comparable interface, which is the standard way for objects to implement a natural ordering.
In this case, String implements Comparable; it uses lexicographic order (which is roughly the order of words in a dictionary), comparing characters pairwise until it finds a difference, or until one String ends. So for example "a" < "abc" < "b".
Collection ordering in Kotlin is explained here.
Have a look at this docu:
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.collections/min.html
Since String implements comparable, min will return the smallest value based on alpha-numeric sorting.
For eval. array by string length:
list.minBy { it -> it.length };
Hi can someone show me how to do this:
Your function should take a Maple list of complex numbers as its input and return the largest modulus from that list.
Hint: Use the map command to apply the abs command to some such list. Then apply the max command to that. Now repeat, by composing that second operation around the first. Finally, create an operator which takes L to that composed operation applied to L.
Look to the help-pages ?max, ?map, ?abs, and ?operators,functional, if necessary.