Javascript If Another Line

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The Java ternary operator allows you to write an if statement on a line of code. A ternary operator can return true or false. Returns a specified value depending on whether the statement is true or false.

We use Java if ... else instructions to control the flow of ’a program. An if statement will evaluate whether an expression is true or false. This declaration executes a specific block of code if the expression equals true.

However, if ... else statements on multiple lines. If you are evaluating a basic expression , your syntax may get unnecessarily verbose. This is where the ternary operator comes in. The Java ternary operator is used to simply replace if ... else statements to make the code easier to read.

This tutorial will discuss how to use the Java ternary operator. We will look at an example so that you can learn how to use this operator. Let’s get started!

Java ternary operator

The Java ternary operator allows you to write concise if ... else statements. Ternary statements get their name because they accept three conditions. A ternary operator evaluates whether a statement is true or false and returns a value determined according to the result of the operator

Here is the syntax for a ternary operator in Java:.

The origin of the name "ternary" refers to the way a ternary operator has three parts. Our declaration accepts three operands: the value assigned to variable if the expression is true

  • expressionIsFalse is the value assigned to variable if the expression is false
  • It does The content of the ternary operator must not be assigned to a variable. For example, you can write a ternary operator in a System.out.println () release. see the result of your ternary operator in the java console

    Unlike an "if" , the ternary operator does not accept "other" word - key statement . The ternary declaration uses a colon (:) to represent an "other" condition.

    Let use an example to show this operator in action.

    Java example of a ternary operator

    Suppose you are creating a commercial website. We want people to be able to buy products only if they are 16 years of age or older.

    To verify the age of our customers, we can use a ternary operator. This is more effective than using an "if" statement because a user cannot be under 16 or over 16. Here is an example of a program that would do the job of checking the age of a user:

    Our code evaluates our ternary. Our condition is true, our code returns:.

    All of first, we’ll define a class called EvaluateAge. Next, we declare a variable Java called age. This variable stores the age value of our client. age is assigned the value 22.

    We declare a variable called "Result‚" whose value is equal to the result of our ternary operator. The ternary operator evaluates whether the "age" is equal to or greater than 16 ("age> = 16 ").

    If the expression returns true , operator comes back "This user is over 16 years old. Otherwise, the operator returns This user is under 16. In the last line of our code, we print the message returned by the Result variable.

    If our user’s age was 15, our code would be the following result of Java string:

    indeed, our ternary evaluates false if the user’s age is not equal to or greater than 16 years. We have successfully created a system to check if a user is old enough to use our services.

    When to use ternary Java operators

    Termaru operators should be used if you have a simple "if" statement that you want to appear more concise and in your code . The ternary operator makes your code more readable.

    In our example above, we evaluated the expression. If we wrote the code to assess the age of our user as "if" statement, we would write:

    if is simple, but spanned five lines. Using a ternary declaration, we’ve reduced our if to a single line.

    In general, you should only use ternary statements when the resulting statement is short. Otherwise, write a normal if statement. The purpose of a ternary operator is to make the code more concise and easier to read. Moving to a complex if statement in a ternary operator goes to the counter to this goal

    Both operators Java ternary and conditional operators of the if statement evaluate Boolean expressions. Boolean expressions are those where the only output can be true or false. To learn more about Java Booleans, see our complete guide Java Booleans .

    Conclusion

    The ternary operator is a feature in Java that allows you to write more concise si statements to control the flow of code. These operators are called ternary because they accept three operands.

    In this tutorial, we have covered the basics of Java ternary operators. He also explored how ternary operators deal with Java if statements, as well as examples of each in action.

    Do you want to know more about coding in Java? Check out our Java Learning Guide . You will find a list of the best online courses and expert advice on learning the Java programming language

    Javascript If Another Line exp: Questions

    exp

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

    5 answers

    Carl Meyer 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

    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)
    

    Javascript If Another Line find: Questions

    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

    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

    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
    

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