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 product 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 ` `# Consecutive Product 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` `# Consecutive Product list` `res ` `=` `[test_list[i] ` `*` `test_list[i ` `+` `1` `] ` `for` `i ` `in` `range` `(` `len` `(test_list)` `-` `1` `)]` ` ` `# printing result` `print` `(` `"The computed successive product list is : "` `+` `str` `(res))` |

**Output :**

The original list is : [1, 4, 5, 3, 6] The computed successive product list is : [4, 20, 15, 18]

**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 product.

`# Python3 code to demonstrate ` `# Consecutive Product 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()` `# Consecutive Product list` `res ` `=` `[i ` `*` `j ` `for` `i, j ` `in` `zip` `(test_list[: ` `-` `1` `], test_list[` `1` `:])]` ` ` `# printing result` `print` `(` `"The computed successive product list is : "` `+` `str` `(res))` |

**Output :**

The original list is : [1, 4, 5, 3, 6] The computed successive product list is : [4, 20, 15, 18]

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