Javascript Salary In India

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Joining a tech company like Google is a dream for many. It represents the pinnacle of large tech companies. Google is everywhere in our life, having products created such as the well known Gmail, GSuite and of course Google Search. No wonder so many software engineers aspire to work for the company. Combine that with a software engineer salary from Google which offers a six-figure range .

Compensation is often the top priority for everything software engineer . Every job has to pay its bills, and some jobs do it better than others. With a job at Google, you will have no problem earning a good salary. But the question remains: What salary can you expect to earn at Google

In this guide, we will discuss what salary software engineers can Google expect ?. We’ll explore how many engineers earn, what benefits they get, and how pay differs between different seniority levels.

Google engineer levels

Google, like many large tech companies, uses a level-based pay system . The large companies often use this approach because their hierarchy is broader. Having a scaling system also ensures that all similar levels within the company earn a comparable salary.

To understand how much you can expect to earn, it’s important to know Google’s software engineer levels. There are nine software development job titles from Google, which are as follows:

  1. Software Engineer II (L3 | 0-1 years experience)
  2. Engineer Software III (L4 | 2 years experience)
  3. senior Software Engineer (L5 | Director I equivalent)
  4. Personal software engineer (L6 | Manager II equivalent)
  5. Senior Software Engineer Staff (L7 | senior Manager)
  6. Senior Engineer (L8 | Director)
  7. Distinguished Engineer (L9 | Senior Manager)
  8. Google Fellow (L10 | vice president)
  9. Google Senior Fellow (L11 | first vice president)
  10. Most Google engineers work in the lower end of the hierarchy. According to some former Google software engineers, it is very difficult to get promotion beyond the software engineer level of the staff. This may be due to the shortage of jobs, while the so-called cultivation cream is being encouraged.

    Google salary software engineer

    The average salary you earn as a software engineer at Google will depend on your level of experience and according to level. If you start out as an IT engineer entry level , your average total salary will naturally tend towards the lower end of the range. ladder . As you accumulate experience and earn promotions, you can expect to earn a higher salary

Google is undoubtedly one of the best tech companies in terms of compensation. At lower levels, the salary is equivalent to that offered by many other companies. But when you look at the high levels, it is clear that Google compensates its top workers favorably.

Here are the average salaries you can expect to earn on Google by level. We have also included the estimated calculation of their bonuses and stock options.

Name Level total base Bonus stocks (/ an)
L3 $ 179,000 $ 120,000 $ 20,000 $ 39,000
L4 $ 256,000 $ 150,000 $ 27,000 $ 78,000
L5 $ 343,000 $ 180,000 $ 36,000 $ 126,000
L6 $ 491,000 $ 210,000 $ 60,000 $ 219.00
L7 $ 705,000 $ 264,000 $ 95,000 346,000 $
L8 1,350,000 $ $ 318,000 122000 $ 907000 $

Data from .

salary data is not available for levels L9 to L11. Needless to say, it can be immediately recognized that the salaries earned by Google software engineers are impressive across the board . To help you better understand these compensation plans, here is a breakdown of their components.

  • Base pay refers to the standard salary you earn. This is the initial rate of pay you will earn per year.
  • Bonus. These services are offered to Google software engineers on an annual basis and may vary from employee to employee .
  • Stocks. Stock refers to stock options provided by Google. We ’ll talk more about this in the next section

All of these data points add up to create your compensation package

Options Google Software stock Engineer

The stock options for Google software engineers are varied. Like many other Silicon Valley tech companies, Google software engineers are awarded Restricted Stock Units (RSU) . The company uses them to compensate the stock employees of the company.

RSUs are subject to a four-year acquisition program at Google. Each year, 25 percent of your inventory is acquired (approximately at a rate of 2.08 percent per month). This means that you will unlock 25% of your shares each year.

Google sometimes refers to its UARs as Google Stock Units (GSU) . In addition, the RSU / GSU exercise program may vary depending on the number of actions received. For example, if you earn less than 32 GSU, your shares will accumulate each year. If you earn between 64 and 159 GSU, your shares will accumulate each quarter.

East - that these salaries impress you? Would you believe that there is more to Google’s engineer salary package than cash bonuses and stock options? Well, as a software engineer, you will also be entitled to a number of benefits, some of which are exclusive to Google.

For example, Google offers its employees a subsidized massage chair on site or bed. . The company also provides employees and their families with personal international travel all year round.

In addition, Google software engineers receive $ 500 after the arrival of a new baby. This attempts to help employees take care of house cleaning , diaper bills and other related expenses. On the other advantages are among the major companies tech standard. These can be broken down into a few different categories.

These are the three main benefits offered to Google employees:.

  • Insurance and welfare benefits
  • Transport benefits
  • National competitions , financial and other benefits

insurance and welfare benefits

Advantage description
insurance view -
Pet Insurance -
Sick leave Unlimited
paternity leave 12 weeks
permit paid 15 days to start; increases to 20 days after working for the company for four years and to 25 days after
Maternity leave 18 weeks < / tr>
Gym on site -
free breakfast, lunch and dinner 5 days a week
dental insurance -

Transport advantages

Advantage description
regional transport system -
Shuttle Company -
bike on campus Allows you to navigate to google headquarters and surrounding area

Home, finance and other benefits h3> Laundry on site
Advantage description
Refund of telephone bills -
fertility and adoption assistance -
reimbursement of fees tuition -
Mega Backdoor Roth IRA -
401 (k) Match 50 percent available on all contributions up to $ 9,500; Google will match the greater of: (a) 100 percent of your contributions up to $ 3,000 or (b) 50 percent of your contributions up to $ 9,500
employee discount <. / Td> Range 5 to 15% discount on Google products
Surrogacy support refund up to $ 40,000 in commissions
match for donations -
Pet Friendly Workstation Seat < / td> & mdash;
With all these factors considered to be

, it is clear that the work on Google is leading significant advantages.

Google vs other tech companies

How Google rates for wages Microsoft and wages on Amazon ? That’s a good question. The table below includes comparisons between the compensation plans offered by Google and those offered by Microsoft and Amazon. This will help you determine which company can earn the highest salary:


Google offers one of the most attractive compensation plans in the tech industry. In both engineer base, you will have a large salary. Additionally , as you gain Google experience, your salary and other compensation options like scholarships and bonuses will increase.

But wages are only part of the equation. You will also receive perks like tuition refunds, a favorable 401 (k) policy, and free meals five days a week. If you are considering job for a large tech company , Google is definitely worth considering.

Frequently asked questions

Javascript Salary In India accumulate: Questions

Javascript Salary In India exp: Questions


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)
>>> 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.)


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)


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()
    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:
    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.


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


>>> 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()
    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}))
>>> 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())))
$ 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


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


Answer #3

An alternative:

z = x.copy()


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Comparison Level Google Other company
Google vs Amazon salary Base level $ 179,000 $ 153,000
Google vs Amazon salary Second level < / td> $ 256,000 $ 213,000
Google vs Amazon salary Third level $ 343,000 $ 312,000
Google vs Microsoft compensation base level $ 179,000 < / td> $ 156,000
Google vs Microsoft salary Second level $ 256,000 $ 170,000 < tr> Google vs Microsoft salary Third level $ 343,000 $ 218,000