List Of Javascript Font Families

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Formatting text on a web page is an important part of web design. For example, you might want the headers to appear in a specific font, or for all text on a page to use the serif font.

This is where the CSS font is located: the family comes into play. The CSS font-family property allows you to specify one or more font families that will be used to style the text on a page Web.

This tutorial will cover the basics by referring to examples of CSS font-family and how you can use the property to style text on a web page.

CSS Font Family

There are many aspects involved in styling a font in CSS, from setting a color to size. the police. One of the most important parts of a font style, however, is defining the font family, which is the style used by text on a web page.

Define the font family used by text on a web page, you can use the font-family property. The syntax of the font-family property is as follows:

The font-family property accepts a single font or a list of fonts. If you specify more than one character, you must separate each character with a comma.

The order in which the characters are displayed is the order in which the browser will attempt to apply the characters. The browser will default to the first font in the list that is installed or can be downloaded from the web browser. If this character is not available, the second character will be used, and so on.

It is important to note that the font-family property will select a font for each character in a block of text. Then, if a particular character is not available in a font, the The next font of your font in the family list will be tried, until a suitable character can be found.

When using font-weight, it is best to include at least one last name generic font (such as serif, sans-serif, or monospace). This ensures that if another font in the list is not available, the browser will still be able to select a font for the text on a web page to display.

There are two ways you can use the font-family property. First, you can use the property to specify a custom font for a web page. Second, you can use the property to apply a generic font family to text on a web page. The generic font families available for use with the font-family property are as follows:

There are other generic font families available, which you can read in more detail in Mozilla CSS Font Family Documentation .

Now that we know the basics of the font-family property, we can look at some examples of how it works in action.

Examples of CSS Font Families

A local stamp asked us club, the Seattle Stamp Club, to customize the fonts for parts of their website.

First of all, we were asked to change the font of all the

headers on their "Contacts" page to use the "Open Sans" font. So, for example, the "Contact Us" on the "Contact Us" page of their website should appear in Open Sans. If this font is not available, the generic "sans-serif" family must be used. We could use the following code to do this:

Click ’Image" noscript> ’Image button in the code editor above to see the output of our HTML / CSS code.

In our HTML code, we have specified a header using an

tag which contains the text Contact Us . So in our CSS code, we used the font family to set the font for the

elements on our webpage.

The value we assigned to the property font-family sets the font for all

elements in Open Sans and, if this font is not available, the generic sans-serif family is used. To achieve this we have specified two characters, separated by a comma.

Now suppose we also want to apply a style to all the characters

elements of the page Contacts from the stamp club to use the Times font. If this font is not available, a default serif font should be used. We could use the following code to do this:

Click ’Image in the code editor above to see the output of our HTML / CSS code.

In our example, we have declared a paragraph of text using a

tag . Next, we define the font family of all

tags in our CSS stylesheet to use the Times and serif fonts. By default, the browser will try to display all

tags in the Times font. If this font is not available, the serif font is used.

Conclusion

The CSS font-family property allows you to specify the style of the text of a web page. For example, you can use the font-family property to use the " Arial " font property or the " Italic " generic font for all text on a Web page.

This tutorial explained, by referring to examples, the basics of the CSS font-family property and how you can use it to style text on a web page. You are now ready to start using font family property like an expert web designer !

List Of Javascript Font Families circle: Questions

circle

How to do a scatter plot with empty circles in Python?

2 answers

In Python, with Matplotlib, how can a scatter plot with empty circles be plotted? The goal is to draw empty circles around some of the colored disks already plotted by scatter(), so as to highlight them, ideally without having to redraw the colored circles.

I tried facecolors=None, to no avail.

204

Answer #1

From the documentation for scatter:

Optional kwargs control the Collection properties; in particular:

    edgecolors:
        The string ‘none’ to plot faces with no outlines
    facecolors:
        The string ‘none’ to plot unfilled outlines

Try the following:

import matplotlib.pyplot as plt 
import numpy as np 

x = np.random.randn(60) 
y = np.random.randn(60)

plt.scatter(x, y, s=80, facecolors="none", edgecolors="r")
plt.show()

example image

Note: For other types of plots see this post on the use of markeredgecolor and markerfacecolor.

circle

plot a circle with pyplot

2 answers

surprisingly I didn"t find a straight-forward description on how to draw a circle with matplotlib.pyplot (please no pylab) taking as input center (x,y) and radius r. I tried some variants of this:

import matplotlib.pyplot as plt
circle=plt.Circle((0,0),2)
# here must be something like circle.plot() or not?
plt.show()

... but still didn"t get it working.

199

Answer #1

You need to add it to an axes. A Circle is a subclass of an Patch, and an axes has an add_patch method. (You can also use add_artist but it"s not recommended.)

Here"s an example of doing this:

import matplotlib.pyplot as plt

circle1 = plt.Circle((0, 0), 0.2, color="r")
circle2 = plt.Circle((0.5, 0.5), 0.2, color="blue")
circle3 = plt.Circle((1, 1), 0.2, color="g", clip_on=False)

fig, ax = plt.subplots() # note we must use plt.subplots, not plt.subplot
# (or if you have an existing figure)
# fig = plt.gcf()
# ax = fig.gca()

ax.add_patch(circle1)
ax.add_patch(circle2)
ax.add_patch(circle3)

fig.savefig("plotcircles.png")

This results in the following figure:

The first circle is at the origin, but by default clip_on is True, so the circle is clipped when ever it extends beyond the axes. The third (green) circle shows what happens when you don"t clip the Artist. It extends beyond the axes (but not beyond the figure, ie the figure size is not automatically adjusted to plot all of your artists).

The units for x, y and radius correspond to data units by default. In this case, I didn"t plot anything on my axes (fig.gca() returns the current axes), and since the limits have never been set, they defaults to an x and y range from 0 to 1.

Here"s a continuation of the example, showing how units matter:

circle1 = plt.Circle((0, 0), 2, color="r")
# now make a circle with no fill, which is good for hi-lighting key results
circle2 = plt.Circle((5, 5), 0.5, color="b", fill=False)
circle3 = plt.Circle((10, 10), 2, color="g", clip_on=False)
    
ax = plt.gca()
ax.cla() # clear things for fresh plot

# change default range so that new circles will work
ax.set_xlim((0, 10))
ax.set_ylim((0, 10))
# some data
ax.plot(range(11), "o", color="black")
# key data point that we are encircling
ax.plot((5), (5), "o", color="y")
    
ax.add_patch(circle1)
ax.add_patch(circle2)
ax.add_patch(circle3)
fig.savefig("plotcircles2.png")

which results in:

You can see how I set the fill of the 2nd circle to False, which is useful for encircling key results (like my yellow data point).

List Of Javascript Font Families 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)

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