# Calculate The Average Javascript

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There are two ways to find the mean of a list of numbers in Python. You can divide the sum () by the length () of a list of numbers to find the mean . Or you can find the mean of a list using the Python function mean ().

Finding the mean of a set of values ‚Äã‚Äãis a common Python task.

Announcement For example, you may have a list of product sales and want to know the average purchase price. Or, for a diversity report, you can calculate the average age of employees working for your organization.

## Python average

There are two methods to find the average of a list of numbers in Python:

• Calculate the sum of the list, then divide that number by the length of the list. The length of the list is the number of values ‚Äã‚Äãin the list. or ;
• Using the statistics. mean () method.

Although both methods return the average of a list of numbers, there are several factors to consider when choosing which one to use.

In this tutorial, we explain how to use the approaches mentioned above to find the mean of a list in Python. We’ll take a look at two examples to get you started.

## Python Average: Len () and Sum ()

The formula for averaging a list of values ‚Äã‚Äãis the sum of all the terms divided by the number of those terms . You can use the Python values ‚Äã‚Äãsum () and len () to calculate the average of numbers in a list.

The Python len () method calculates and returns a number of numbers in a list. len is the abbreviation for length, which refers to the length or number of elements in a list.

The sum () in Python calculates the sum total of all values ‚Äã‚Äãof a list.

We can use these functions together to calculate the mean value from ’a list of values.

The basic formula to calculate an average in Python is:

This approach is common because you don’t need to import external values ‚Äã‚Äãto calculate an average.

### Sum () and len () function example

Let’s say we have a cafe and want to know the average price of all the orders placed to date. To calculate this number, we could use the sum () and len () methods we just saw:

We print the average value on the console:

In the first line of our code, we define a Python variable . This variable stores an order price list. Next, we calculate the average of these orders by dividing the sum of all the orders by the length of our list of orders.

At the end of our code, we print a message to the console telling us the average price of coffee orders.

We use the Python round () method to round our average to two decimal places. The str () method converts our average value to a string that we can print to the console.

We now know that the average price for a coffee order to date is \$ 3.69.

## Python average: statistics. mean ()

The Python statistics library contains a function called statistics. mean () which calculates the average value of a list of values. The term mean is used in mathematics to describe the mean of a list of values.

The syntax of the statistics. mean () method is:

statistics. mean (list_of_values)

The media () takes one parameter: the list of items you want to average.

Before using this method, you need to import the statistics module (statistics) into Python This is a built-in module that can be used to perform various calculations in Python.

This is an important difference between using the statistics. mean () and sum () and len () methods. sum () and len () can be used without importing a library. However, you must import statistics to use statistics. mean ().

If you don’t mind importing another library, this method works fine. Since you can calculate an average value without any libraries, you should consider rare to do so before using the statistics library.

We can import the statistics module using a Python import statement :

We can now access the Mean () in the statistics library.

### statistic example. mean ()

We are now ready to start calculating averages using Mean ().

Let’s take the example of the coffee orders we used above. Suppose we want to calculate the average price of coffee orders placed to date. We could use the following code to calculate the average price:

Let’s take a look at the output of our program:

Our program returns the same text and output as our first example.

### Calculate the average of a tuple

We can use statistics. mean () to calculate the average value of a tuple, which is a sequence orderly and immutable elements. Here is an example of statistics. mean () used to calculate the average price of a coffee order, with our information stored in a tuple:

Our code returns the same response as in the last example because we let’s work with the same numbers:

A tuple can be used in place of a list if you are working with values ‚Äã‚Äãthat should not change. So, since the prices of our coffee orders will not change after an order is processed, a tuple is appropriate.

### Average of negative values ‚Äã‚Äã

You can use average () to calculate the average value of a set of negative numbers.

Suppose we have a list that stores the lowest daily temperatures recorded during a specific week in winter. We could use the following code to calculate the average of the lowest daily temperatures for that week:

Our code returns:

## Conclusion

< p> You can calculate the average of a list of values ‚Äã‚Äãusing the sum () and the len () or the statistics. mean () method. The statistics. mean () method must be imported into your program. You can use the sum () and len () methods without importing any external libraries into your code.

Want to learn more about programming in Python ? Read our Python Learning Guide . This guide contains a list of the best learning resources you can use to improve your knowledge.

👻 Read also: what is the best laptop for engineering students?

## Calculate The Average Javascript exp: Questions

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

By Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The `update()` method would be what I need, if it returned its result instead of modifying a dictionary in-place.

``````>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}
``````

How can I get that final merged dictionary in `z`, not `x`?

(To be extra-clear, the last-one-wins conflict-handling of `dict.update()` is what I"m looking for as well.)

5839

## How can I merge two Python dictionaries in a single expression?

For dictionaries `x` and `y`, `z` becomes a shallowly-merged dictionary with values from `y` replacing those from `x`.

• In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

``````z = x | y          # NOTE: 3.9+ ONLY
``````
• In Python 3.5 or greater:

``````z = {**x, **y}
``````
• In Python 2, (or 3.4 or lower) write a function:

``````def merge_two_dicts(x, y):
z.update(y)    # modifies z with keys and values of y
return z
``````

and now:

``````z = merge_two_dicts(x, y)
``````

### Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

``````x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}
``````

The desired result is to get a new dictionary (`z`) with the values merged, and the second dictionary"s values overwriting those from the first.

``````>>> z
{"a": 1, "b": 3, "c": 4}
``````

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

``````z = {**x, **y}
``````

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

``````z = {**x, "foo": 1, "bar": 2, **y}
``````

and now:

``````>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}
``````

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

``````z = x.copy()
z.update(y) # which returns None since it mutates z
``````

In both approaches, `y` will come second and its values will replace `x`"s values, thus `b` will point to `3` in our final result.

## Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

``````def merge_two_dicts(x, y):
"""Given two dictionaries, merge them into a new dict as a shallow copy."""
z = x.copy()
z.update(y)
return z
``````

and then you have a single expression:

``````z = merge_two_dicts(x, y)
``````

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

``````def merge_dicts(*dict_args):
"""
Given any number of dictionaries, shallow copy and merge into a new dict,
precedence goes to key-value pairs in latter dictionaries.
"""
result = {}
for dictionary in dict_args:
result.update(dictionary)
return result
``````

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries `a` to `g`:

``````z = merge_dicts(a, b, c, d, e, f, g)
``````

and key-value pairs in `g` will take precedence over dictionaries `a` to `f`, and so on.

Don"t use what you see in the formerly accepted answer:

``````z = dict(x.items() + y.items())
``````

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two `dict_items` objects together, not two lists -

``````>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"
``````

and you would have to explicitly create them as lists, e.g. `z = dict(list(x.items()) + list(y.items()))`. This is a waste of resources and computation power.

Similarly, taking the union of `items()` in Python 3 (`viewitems()` in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

``````>>> c = dict(a.items() | b.items())
``````

This example demonstrates what happens when values are unhashable:

``````>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"
``````

Here"s an example where `y` should have precedence, but instead the value from `x` is retained due to the arbitrary order of sets:

``````>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}
``````

Another hack you should not use:

``````z = dict(x, **y)
``````

This uses the `dict` constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. `frozenset`s or tuples), but this method fails in Python 3 when keys are not strings.

``````>>> c = dict(a, **b)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings
``````

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for `dict(**y)` is for creating dictionaries for readability purposes, e.g.:

``````dict(a=1, b=10, c=11)
``````

``````{"a": 1, "b": 10, "c": 11}
``````

Despite what Guido says, `dict(x, **y)` is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. `dict` broke this consistency in Python 2:

``````>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}
``````

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

`dict(x.items() + y.items())` is still the most readable solution for Python 2. Readability counts.

My response: `merge_two_dicts(x, y)` actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

`{**x, **y}` does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

``````from copy import deepcopy

def dict_of_dicts_merge(x, y):
z = {}
overlapping_keys = x.keys() & y.keys()
for key in overlapping_keys:
z[key] = dict_of_dicts_merge(x[key], y[key])
for key in x.keys() - overlapping_keys:
z[key] = deepcopy(x[key])
for key in y.keys() - overlapping_keys:
z[key] = deepcopy(y[key])
return z
``````

Usage:

``````>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}
``````

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

## Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than `copy` and `update` or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

``````{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7
``````

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

``````dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2
``````

`itertools.chain` will chain the iterators over the key-value pairs in the correct order:

``````from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2
``````

## Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

``````from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
z = x.copy()
z.update(y)
return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
``````

In Python 3.8.1, NixOS:

``````>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
``````
``````\$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux
``````

## Resources on Dictionaries

5839

In your case, what you can do is:

``````z = dict(list(x.items()) + list(y.items()))
``````

This will, as you want it, put the final dict in `z`, and make the value for key `b` be properly overridden by the second (`y`) dict"s value:

``````>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

``````

If you use Python 2, you can even remove the `list()` calls. To create z:

``````>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}
``````

If you use Python version 3.9.0a4 or greater, then you can directly use:

``````x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
``````
``````{"a": 1, "c": 11, "b": 10}
``````

5839

An alternative:

``````z = x.copy()
z.update(y)
``````

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 Calculate The Average Javascript, check other exp-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:

Dmitry Schteiner

California | 2022-12-01

Maybe there are another answers? What Calculate The Average Javascript exactly means?. Will get back tomorrow with feedback

Angelo Krasiko

Berlin | 2022-12-01

find is always a bit confusing 😭 Calculate The Average Javascript is not the only problem I encountered. I am just not quite sure it is the best method

Olivia Robinson

Paris | 2022-12-01

exp is always a bit confusing 😭 Calculate The Average Javascript is not the only problem I encountered. Will use it in my bachelor thesis

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