Difference between consecutive elements in list Python

Python | Generate successive element difference list

While working with python, we usually come by many problems that we need to solve in day-day and in development. Specially in development, small tasks of python are desired to be performed in just one line. We discuss some ways to compute a list consisting of elements that are successive difference in the list.

Method #1 : Using list comprehension
Naive method can be used to perform, but as this article discusses the one liner solutions to this particular problem, we start with the list comprehension as a method to perform this task.




# Python3 code to demonstrate
# to generate successive difference list
# using list comprehension
# initializing list
test_list = [1, 4, 5, 3, 6]
# printing original list
print ("The original list is : " + str(test_list))
# using list comprehension
# generate successive difference list
res = [test_list[i + 1] - test_list[i] for i in range(len(test_list)-1)]
# printing result
print ("The computed successive difference list is : " + str(res))

Output :

The original list is : [1, 4, 5, 3, 6] The computed successive difference list is : [3, 1, -2, 3]

Method #2 : Using zip()
zip() can also be used to perform the similar task and uses the power of negative indices to zip() the index element with its next element and hence compute the difference.




# Python3 code to demonstrate
# to generate successive difference list
# using zip()
# initializing list
test_list = [1, 4, 5, 3, 6]
# printing original list
print ("The original list is : " + str(test_list))
# using zip()
# generate successive difference list
res = [j - i for i, j in zip(test_list[: -1], test_list[1 :])]
# printing result
print ("The computed successive difference list is : " + str(res))

Output :

The original list is : [1, 4, 5, 3, 6] The computed successive difference list is : [3, 1, -2, 3]

Method #3 : Using map() + operator.sub
map() can be coupled with the subtraction operator to perform this particular task. This maps the element with its next element and performs the subtraction operation. Other operators can be passed to perform the desired operations.




# Python3 code to demonstrate
# to generate successive difference list
# using map() + operator.sub
import operator
# initializing list
test_list = [1, 4, 5, 3, 6]
# printing original list
print ("The original list is : " + str(test_list))
# using map() + operator.sub
# generate successive difference list
res = list(map(operator.sub, test_list[1:], test_list[:-1]))
# printing result
print ("The computed successive difference list is : " + str(res))

Output :

The original list is : [1, 4, 5, 3, 6] The computed successive difference list is : [3, 1, -2, 3]




Article Tags :
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Python list-programs
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Practice Tags :
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Python: Find the difference between consecutive numbers in a given list

Last update on December 10 2020 11:24:43 (UTC/GMT +8 hours)

numpy.ediff1d¶

numpy.ediff1d(ary, to_end=None, to_begin=None)[source]

The differences between consecutive elements of an array.

Parametersaryarray_like

If necessary, will be flattened before the differences are taken.

to_endarray_like, optional

Number(s) to append at the end of the returned differences.

to_beginarray_like, optional

Number(s) to prepend at the beginning of the returned differences.

Returnsediff1dndarray

The differences. Loosely, this is ary.flat[1:] - ary.flat[:-1].

See also

diff, gradient

Notes

When applied to masked arrays, this function drops the mask information if the to_begin and/or to_end parameters are used.

Examples

>>> x = np.array([1, 2, 4, 7, 0]) >>> np.ediff1d(x) array([ 1, 2, 3, -7])
>>> np.ediff1d(x, to_begin=-99, to_end=np.array([88, 99])) array([-99, 1, 2, ..., -7, 88, 99])

The returned array is always 1D.

>>> y = [[1, 2, 4], [1, 6, 24]] >>> np.ediff1d(y) array([ 1, 2, -3, 5, 18])

Compute the differences between consecutive elements and append a number in Numpy

NumpyServer Side ProgrammingProgramming

To compute the differences between consecutive elements of a masked array, use the MaskedArray.ediff1d() method in Python Numpy. The "to_end" parameter sets the number(s) to append at the end of the returned differences.

This function is the equivalent of numpy.ediff1d that takes masked values into account, see numpy.ediff1d for details.

A masked array is the combination of a standard numpy.ndarray and a mask. A mask is either nomask, indicating that no value of the associated array is invalid, or an array of booleans that determines for each element of the associated array whether the value is valid or not.

C++

// C++ program to maximize the sum of difference
// between consecutive elements in circular array
#include <bits/stdc++.h>
using namespace std;
// Return the maximum Sum of difference between
// consecutive elements.
int maxSum(int arr[],int n)
{
int sum = 0;
// Sorting the array.
sort(arr, arr + n);
// Subtracting a1, a2, a3,....., a(n/2)-1, an/2
// twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.
// ...., an - 1, an twice.
for (int i = 0; i < n/2; i++)
{
sum -= (2 * arr[i]);
sum += (2 * arr[n - i - 1]);
}
return sum;
}
// Driver Program
int main()
{
int arr[] = { 4, 2, 1, 8 };
int n =sizeof(arr)/sizeof(arr[0]);
cout << maxSum(arr, n) << endl;
return 0;
}

Java

// Java program to maximize the sum of difference
// between consecutive elements in circular array
import java.io.*;
import java.util.Arrays;
class MaxSum
{
// Return the maximum Sum of difference between
// consecutive elements.
static int maxSum(int arr[],int n)
{
int sum =0;
// Sorting the array.
Arrays.sort(arr);
// Subtracting a1, a2, a3,....., a(n/2)-1,
// an/2 twice and adding a(n/2)+1, a(n/2)+2,
// a(n/2)+3,....., an - 1, an twice.
for (int i =0; i < n/2; i++)
{
sum -= (2 * arr[i]);
sum += (2 * arr[n - i -1]);
}
return sum;
}
// Driver Program
public static void main (String[] args)
{
int arr[] = {4,2,1,8 };
int n = arr.length;
System.out.println(maxSum(arr, n));
}
}
/*This code is contributed by Prakriti Gupta*/
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