# Javascript Multiplies The String By A Number

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You can multiply two numbers together in Python . You can also multiply a number by a string. This returns a sequence of a string that repeats a specific number of times.

If you try to multiply one string by another string, you will encounter the error " TypeError: unable to multiply sequence for non -int of type ’str’" error.

In this guide, we explain what this error means and where you might encounter it in your code. Let’s look at an example of this error to help you understand how to fix it.

## TypeError: Cannot multiply sequence by non-int of type ’str’

Strings are a type of sequence . This is because they contain characters that Python can iterate over. Other types of sequences include tuples, dictionaries, and lists.

Use the multiplication operator (*) to create a string that repeats the contents of a string. Consider the following code:

This code returns: "Cakes! Cakes! ". The multiplication operator causes our string to repeat twice.

You cannot use the multiplication operator to multiply a string by a string. Integers and numbers to Floating point are the only values ‚Äã‚Äãthat can be multiplied by values ‚Äã‚Äãof the same data type. Python has no way of interpreting the multiplication of two strings.

## An example scenario

Let’s create a program that calculates how much money a restaurant made from their jam scones on a Thursday afternoon.

To begin with, we need to set the price of a jam scone. We also ask that you tell us how many jam scones you sold in the last day of operation using an ’ input () :

Next, let’s multiply those two values ‚Äã‚Äãtogether. This will tell us how much money was made with jam scones on a given day:

Now that we know how much the scones have won, let’s write a `print ()` statement that informs the user of the result of our calculation:

The format () method allows us to add the value of "earned_from_scones" where our braces ({}) appear in our string.

We also need to calculate how much profit is made on each muffin. We know we make 25 cents on every muffin. To calculate how much profit is made, we use this code:

This code calculates the profit made on each muffin, then prints that value to the console. Let’s run our code and see if it works:

Our code returns an error.

Let’s analyze the line of code of our error:

Although this li gne code looks correct, there is a problem: we are trying to multiply two string values ‚Äã‚Äãtogether. We store "jam_scone" as a string. The `input ()` method returns a string, which means that `sold ()` has a string value.

To solve this problem, we need to make sure that " jam_scone " and " sold " are floating points. This allows us to perform a mathematical operation on these values.

Next, let’s change the way we declare the values ‚Äã‚Äãof these variables:

The value of" jam_scone " is no longer in quotes . This shows that we have changed "jam_scone" from a string to a float. We also used the float () method to convert the "sold" value to a floating point number.

Run our code and see what happens:

Our code is working fine. First, our code asks the user to enter how many scones are sold in the console. Then our program calculates how much money the store has made selling scones. Finally, our program calculates the profits generated by sales of scone.

## Conclusion

The error " TypeError: cannot multiply the sequence by the non-integer of type ’str’ ’occurs if you try to multiply the values ‚Äã‚Äãof two strings together. You can solve this problem by making sure to multiply two numeric values ‚Äã‚Äãtogether or just multiply a string by an integer.

You are now ready to resolve this error as a Pythonist!

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

## Javascript Multiplies The String By A Number __dict__: 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)
``````

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)
``````

We hope this article has helped you to resolve the problem. Apart from Javascript Multiplies The String By A Number, check other __dict__-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:

Angelo Innsbruck

Paris | 2022-11-30

Maybe there are another answers? What Javascript Multiplies The String By A Number exactly means?. Will use it in my bachelor thesis

Chen Danburry

Warsaw | 2022-11-30

Thanks for explaining! I was stuck with Javascript Multiplies The String By A Number for some hours, finally got it done 🤗. Checked yesterday, it works!

Boris Zelotti

Munchen | 2022-11-30

JavaScript is always a bit confusing 😭 Javascript Multiplies The String By A Number is not the only problem I encountered. I just hope that will not emerge anymore

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