## How to get first AND last element of tuple at the same time

python tuple

last item of tuple python

get second element of tuple in list

python get nth element of list of lists

what is the syntax to obtain the first element of the tuple:

remove element from tuple python

how to get value from list of tuples

I need to get the first and last dimension of an numpy.ndarray of arbitrary size.

If I have `shape(A) = (3,4,4,4,4,4,4,3)`

my first Idea would be to do `result = shape(A)[0,-1]`

but that doesn't seem to work with tuples, why not ??

Is there a neater way of doing this than

s=shape(A) result=(s[0], s[-1])

Thanks for any help

I don't know what's wrong about

(s[0], s[-1])

A different option is to use `operator.itemgetter()`

:

from operator import itemgetter itemgetter(0, -1)(s)

I don't think this is any better, though. (It might be slightly faster if you don't count the time needed to instantiate the `itemgetter`

instance, which can be reused if this operation is needed often.)

**Introduction to the Art of Programming Using Scala,** The approach here is to make a tuple that contains the original elements first and a tuple and the first and last element appear only once in the first and last tuples. functions that need to go through two or three collections at the same time. Accessing the first element and the last element of the tuple requires index operator([]). You can find the first element and last element of tuple using Python. Get your single element of tuple with the methods given here. In addition to this, you can also get elements from first to the end of given tuple using Python.

s = (3,4,4,4,4,4,4,3) result = s[0], s[-1]

**Introduction to Programming and Problem-Solving Using Scala,** The approach here is to make a tuple that contains the original elements first and a tuple and the first and last element appear only once in the first and last tuples. functions that need to go through two or three collections at the same time. Modify / Replace the element at specific index in tuple. To replace the element at index n in tuple we will use the same slicing logic as above, but we will slice the tuple from from (0 to n-1) and (n+1 to end) i.e. # Sliced copy containing elements from 0 to n-1 tupleObj[ : n] # Sliced copy containing elements from n to end tupleObj[n + 1 : ]

If you are using numpy array, then you may do that

s = numpy.array([3,4,4,4,4,4,4,3]) result = s[[0,-1]]

where `[0,-1]`

is the index of the first and last element. It also allow more complex extraction such as `s[2:4]`

**Python program to interchange first and last elements in a list ,** Approach #3: Swap the first and last element is using tuple variable. Store the first and last element as a pair in a tuple variable, say get, and unpack those Tuples can be indexed just like lists. The main difference between tuples and lists is that tuples are immutable - you can't set the elements of a tuple to different values, or add or remove elements like you can from a list. But other than that, in most situations, they work pretty much the same.

Came across this late; however, just to add a non indexed approach as previously mentioned.

Tuple unpacking can be easily be applied to acquire the first and last elements. Note: The below code is utilizing the special asterisk '*' syntax which returns a list of the middle section, having a and c storing the first and last values.

Ex.

A= (3,4,4,4,4,4,4,3) a, *b, c = A print((a, c))

Output (3, 3)

**Python,** Recommended Posts: Python | Find the tuples containing the given element from a list of tuples · Python | Get first element with maximum value in list of tuples I'm trying to obtain the n-th elements from a list of tuples. I have something like: elements = [(1,1,1),(2,3,7),(3,5,10)] I wish to extract only the second elements of each tuple into a list: seconds = [1, 3, 5] I know that it could be done with a for loop but I wanted to know if there's another way since I have thousands of tuples.

**get the last element from a tuple in Python,** Building starts with Ground (0) floor followed by 1st, 2nd floor and so on. Similar is the concept of indexing. Indexing is accessing elements from sequence. Access Tuple Elements. There are various ways in which we can access the elements of a tuple. 1. Indexing. We can use the index operator [] to access an item in a tuple, where the index starts from 0. So, a tuple having 6 elements will have indices from 0 to 5.

**Mathematical Methods in Interdisciplinary Sciences,** (fff;..so), , (zi,z, ,x)| then the elements of i-tuple of A, are added at the end of i-tuple of Al if the elements of these tuples belong to the same set; otherwise, the rule for u) It means that if we go through elements of ā' = (a,a), aft) from its first element as to its last element as and At the same time, a; = 3 as well as a;, = 3. Python 3 only answer (that doesn't use slicing or throw away the rest of the list, but might be good enough anyway) is use unpacking generalizations to get first and last separate from the middle: first, *_, last = some_list

**Python: Get an item of a tuple,** Python Exercises, Practice and Solution: Write a Python program to get the 4th element and 4th element from last of a tuple. A little Linq will do the trick: var myStringList = myTupleList.Select(t=>t.Item1).ToList(); As an explanation, since Tim posted pretty much the same answer, Select() creates a 1:1 "projection"; it takes each input element of the Enumerable, and for each of them it evaluates the lambda expression and returns the result as an element of a new Enumerable having the same number of elements.

##### Comments

- I was just wondering, because I haven't worked with tuples that much so far. Thanks, I'm going to accept your answer as soon as I can (~8 minutes :))
- yes, but shape does not return a np.array. The strange thing is, that the
`s[2:4]`

access even is possible for a tuple, but`s[0, -1]`

is not