# Javascript Median Function

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Python Median ():
With the Python Statistics module, you can find the mean -median-mode-in-python-without-libraries/">median, or mean value, of a data set. The Python mean -median-mode-in-python-without-libraries/">median () function allows you to calculate the mean -median-mode-in-python-without-libraries/">median of any data set without sorting the list first.

When working with numbers in Python, you may want to calculate the mean -median-mode-in-python-without-libraries/">median of a list of numbers . For example, suppose you are creating a program to obtain information about the age of students in a grade four class. You may want to calculate the average age of the students.

This is where the `statistics.mean -median-mode-in-python-without-libraries/">median ()` function comes in. The `statistics.mean -median-mode-in-python-without-libraries/">median () function is part of the Python module statistics and allows you to find the mean -median-mode-in-python-without-libraries/">median of a list of values.`

` This tutorial will explain how to use statistic .mean -median-mode-in-python-without-libraries/">median () method. We will also look at an example of statistics.mean -median-mode-in-python-without-libraries/">median () in action and analyze, line by line, how the method works. This tutorial will explain how to use the statistical method .mean -median-mode-in-python-without-libraries/">median (). We’ll also take a look at an example of statistics.mean -median-mode-in-python-without-libraries/">median () in action and analyze how the method works. Python MedianIn statistics, the mean -median-mode-in-python-without-libraries/">median is the medium value in an ordered list of numbers. For example, for a dataset with the numbers 9, 3, 6, 1, and 4, the mean -median-mode-in-python-without-libraries/">median is 4. When parsing and describing a dataset, the mean -median-mode-in-python-without-libraries/">median is often used with mean , standard deviation, and other statistical calculations. In Python, the statistics.mean -median-mode-in-python-without-libraries/">median () function is used to calculate the mean -median-mode-in-python-without-libraries/">median value of a dataset. `
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` statistics. mean -median-mode-in-python-without-libraries/">median () is part of the statistics for Python module . It includes a number of functions for statistical analysis. First, import the statistics module with this code: We can now use the mean -median-mode-in-python-without-libraries/">median () function. Here is the syntax of the statistics.mean -median-mode-in-python-without-libraries/">median () method: The mean -median-mode-in-python-without-libraries/">median () method accepts one parameter: the list of data. `
` When you call the mean -median-mode-in-python-without-libraries/">median () method, it will sort a list of values ‚Äã‚Äãand find its mean value.If the number of data points is odd, the central data point. If the number is even, the mean -median-mode-in-python-without-libraries/">median is the midpoint between the two central values. Python Median ExampleLet’s take an example. Suppose we build a program that compute the age of all students in a fourth year class to know their age distribution. We want to use mean -median-mode-in-python-without-libraries/">median () to find out the average age of the class. We could use this code: Our program returns: 9.5 This is the average age of the students. Let’s break down our code. On the first line, we import the statistics module to access the mean -median-mode-in-python-without-libraries/">median () method in our code. So, let’s define a list. The list is called student_ages . Stores the ages of all students. Next, we use mean -median-mode-in-python-without-libraries/">median () and pass the student_ages variable to calculate the average age. We assign the returned value to the mean -median-mode-in-python-without-libraries/">median_age variable. Then we use print () to print the value of mean -median-mode-in-python-without-libraries/">median_age to the console. When working with mean -median-mode-in-python-without-libraries/">median (), the list you use must have at least one value. If you pass an empty list via the mean -median-mode-in-python-without-libraries/">median () method, you get this error: Median of a tupleYou can also use the mean -median-mode-in-python-without-libraries/">median method to find the mean -median-mode-in-python-without-libraries/">median of a Tuple Python . A tuple is an ordered and immutable data type. This mean s that it is useful when you want to store similar data that will not change over time. Tuples are declared as a list of values ‚Äã‚Äãseparated by commas and surrounded by braces. `
` Suppose our student age dataset is stored as a tuple rather than a listing. If we wanted to calculate the average age of fourth graders, we could do so using the mean -median-mode-in-python-without-libraries/">median () method. Here is the code we would use to calculate the mean -median-mode-in-python-without-libraries/">median of our tuple: Our code returns: 9.5 . The mean -median-mode-in-python-without-libraries/">median is the same, but instead of storing our data in a list, it’s stored in a tuple. This is obvious because our data is in parentheses (() ) instead of brackets ([]). ConclusionYou can use the statistics.mean -median-mode-in-python-without-libraries/">median () method to calculate the mean -median-mode-in-python-without-libraries/">median of a dataset. Using examples, this tutorial demonstrated how to use the statistics.mean -median-mode-in-python-without-libraries/">median () method in Python to calculate the average value of a list and a tuple. You now know that you should start using the statistics.mean -median-mode-in-python-without-libraries/">median () method like a professional Python developer! `
``` 👻 Read also: what is the best laptop for engineering students? Javascript Median Function exp: Questions exp How do I merge two dictionaries in a single expression (taking union of dictionaries)? 5 answers 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 Answer #1 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 = x.copy() # start with keys and values of x 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. Critiques of Other Answers 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. frozensets 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) instead of {"a": 1, "b": 10, "c": 11} Response to comments 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. More comments: 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 My explanation of Python"s dictionary implementation, updated for 3.6. Answer on how to add new keys to a dictionary Mapping two lists into a dictionary The official Python docs on dictionaries The Dictionary Even Mightier - talk by Brandon Rhodes at Pycon 2017 Modern Python Dictionaries, A Confluence of Great Ideas - talk by Raymond Hettinger at Pycon 2017 5839 Answer #2 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 Answer #3 An alternative: z = x.copy() z.update(y) find Finding the index of an item in a list 5 answers Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python? 3740 Answer #1 >>> ["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 Check for it first with item in my_list (clean, readable approach), or 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 Answer #2 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 Answer #3 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 Javascript Median Function, 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: Italiano Javascript Median FunctionDeutsch Javascript Median FunctionFrançais Javascript Median FunctionEspañol Javascript Median FunctionTürk Javascript Median FunctionРусский Javascript Median FunctionPortuguês Javascript Median FunctionPolski Javascript Median FunctionNederlandse Javascript Median Function中文 Javascript Median Function한국어 Javascript Median Function日本語 Javascript Median Functionहिन्दी Javascript Median Function Carlo Sikorski Prague | 2022-11-30 Thanks for explaining! I was stuck with Javascript Median Function for some hours, finally got it done 🤗. I just hope that will not emerge anymore Cornwall Zelotti Prague | 2022-11-30 Thanks for explaining! I was stuck with Javascript Median Function for some hours, finally got it done 🤗. Checked yesterday, it works! Jan Williams Texas | 2022-11-30 I was preparing for my coding interview, thanks for clarifying this - Javascript Median Function in Python is not the simplest one. 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