# Python | Find the maximum difference between pairs of tuples

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Method # 1: Using ` max () ` + List Comprehension
A combination of these functions can be used to accomplish this task. In this we calculate the absolute difference of all pairs and then return the maximum using ` max () `.

` `

 ` # Python3 demo code work ` ` Maximum difference between a pair of tuples ` ` # Using list comprehension + max () ` ` # initialize the list ` ` test_list ` ` = ` ` [(` ` 3 ` `, ` ` 5 ` `), (` ` 1 ` `, ` ` 7 ` `), (` ` 10 ` `, ` ` 3 ` `), (` ` 1 ` `, ` ` 2 ` `)] ` ` # print original list ` ` print ` ` (` ` "The original list:" ` ` + ` ` str ` ` (test_list)) ` ` ` ` Maximum difference between a pair of tuples ` ` # Using list comprehension + max () ` ` temp ` ` = ` ` [` ` abs ` ` (b ` ` - ` ` a) ` ` for ` ` a, b ` ` in ` ` test_list] ` ` res ` ` = ` ` max ` ` (temp) ` ` # print result ` ` print ` ` (` ` "Maximum difference among pairs: "` ` + ` ` str ` ` (res)) `
` `

` ` Output:

` The original list: [(3, 5) , (1, 7), (10, 3), (1, 2)] Maximum difference among pairs: 7 `

Method # 2: Using ` max () ` + lambda
This is similar to the above method. In this case, the task performed by comprehending the list is solved using a lambda function that provides the logic for calculating the differences. Returns the max. pair difference.

 ` # Python3 code to demonstrate how it works ` ` Maximum difference between a pair of tuples ` ` # Using lambda + max () ` ` # initialize the list ` ` test_list ` ` = ` ` [(` ` 3 ` `, ` ` 5 ` `), (` ` 1 ` `, ` ` 7 ` `), (` ` 10 ` `, ` ` 3 ` `), (` ` 1 ` `, ` ` 2 ` `)] ` ` # print original list ` ` print ` ` (` ` "The original list:" ` ` + ` ` str ` ` (test_list)) ` ` ` ` Maximum difference of a pair of tuples ` ` # Using lambda + max () ` ` res ` ` = ` ` max ` ` (test_list, key ` ` = ` ` lambda ` ` sub: ` ` abs ` ` (sub [` ` 1 ` ` ] ` ` - ` ` sub [` ` 0 ` `])) ` ` # print result ` ` print ` ` (` ` "Maximum difference among pairs:" ` ` + ` ` str ` ` (res)) `

Output:

` The original list: [(3, 5), (1, 7), (10, 3), (1, 2)] Maximum difference among pairs: (10, 3) `

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## Python | Find the maximum difference between pairs of tuples absolute: Questions

How to get an absolute file path in Python

By izb

Given a path such as `"mydir/myfile.txt"`, how do I find the file"s absolute path relative to the current working directory in Python? E.g. on Windows, I might end up with:

``````"C:/example/cwd/mydir/myfile.txt"
``````
877

``````>>> import os
>>> os.path.abspath("mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"
``````

Also works if it is already an absolute path:

``````>>> import os
>>> os.path.abspath("C:/example/cwd/mydir/myfile.txt")
"C:/example/cwd/mydir/myfile.txt"
``````

## Python | Find the maximum difference between pairs of tuples absolute: Questions

How to check if a path is absolute path or relative path in a cross-platform way with Python?

UNIX absolute path starts with "/", whereas Windows starts with alphabet "C:" or "". Does python have a standard function to check if a path is absolute or relative?

175

`os.path.isabs` returns `True` if the path is absolute, `False` if not. The documentation says it works in windows (I can confirm it works in Linux personally).

``````os.path.isabs(my_path)
``````

## Python | Find the maximum difference between pairs of tuples absolute: Questions

How to join absolute and relative urls?

I have two urls:

``````url1 = "http://127.0.0.1/test1/test2/test3/test5.xml"
url2 = "../../test4/test6.xml"
``````

How can I get an absolute url for url2?

139

You should use urlparse.urljoin :

``````>>> import urlparse
>>> urlparse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"
``````

With Python 3 (where urlparse is renamed to urllib.parse) you could use it as follow:

``````>>> import urllib.parse
>>> urllib.parse.urljoin(url1, url2)
"http://127.0.0.1/test1/test4/test6.xml"
``````

Finding the index of an item in a list

Given a list `["foo", "bar", "baz"]` and an item in the list `"bar"`, how do I get its index (`1`) in Python?

3740

``````>>> ["foo", "bar", "baz"].index("bar")
1
``````

Reference: Data Structures > More on Lists

# Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, `index` is a rather weak component of the `list` API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about `list.index` follow. It is probably worth initially taking a look at the documentation for it:

``````list.index(x[, start[, end]])
``````

Return zero-based index in the list of the first item whose value is equal to x. Raises a `ValueError` if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

## Linear time-complexity in list length

An `index` call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give `index` a hint. For instance, in this snippet, `l.index(999_999, 999_990, 1_000_000)` is roughly five orders of magnitude faster than straight `l.index(999_999)`, because the former only has to search 10 entries, while the latter searches a million:

``````>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514

``````

## Only returns the index of the first match to its argument

A call to `index` searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

``````>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2
``````

Most places where I once would have used `index`, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for `index`, take a look at these excellent Python features.

## Throws if element not present in list

A call to `index` results in a `ValueError` if the item"s not present.

``````>>> [1, 1].index(2)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: 2 is not in list
``````

If the item might not be present in the list, you should either

1. Check for it first with `item in my_list` (clean, readable approach), or
2. Wrap the `index` call in a `try/except` block which catches `ValueError` (probably faster, at least when the list to search is long, and the item is usually present.)

3740

One thing that is really helpful in learning Python is to use the interactive help function:

``````>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
...

|
|  index(...)
|      L.index(value, [start, [stop]]) -> integer -- return first index of value
|
``````

which will often lead you to the method you are looking for.

3740

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use `enumerate()`:

``````for i, j in enumerate(["foo", "bar", "baz"]):
if j == "bar":
print(i)
``````

The `index()` function only returns the first occurrence, while `enumerate()` returns all occurrences.

As a list comprehension:

``````[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
``````

Here"s also another small solution with `itertools.count()` (which is pretty much the same approach as enumerate):

``````from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
``````

This is more efficient for larger lists than using `enumerate()`:

``````\$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
\$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop
``````

We hope this article has helped you to resolve the problem. Apart from Python | Find the maximum difference between pairs of tuples, check other absolute-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:

Olivia Richtgofen

Rome | 2022-11-27

I was preparing for my coding interview, thanks for clarifying this - Python | Find the maximum difference between pairs of tuples in Python is not the simplest one. Checked yesterday, it works!

Anna Gonzalez

Tallinn | 2022-11-27

Maybe there are another answers? What Python | Find the maximum difference between pairs of tuples exactly means?. I am just not quite sure it is the best method

Anna Williams

Abu Dhabi | 2022-11-27

Maybe there are another answers? What Python | Find the maximum difference between pairs of tuples exactly means?. I am just not quite sure it is the best method

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