Javascript Gets The Child Element By Tag

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To create a CSS bold text effect , you must use the font-weight property. The font-weight property determines the "weight" of a font, or how bold that font appears. You can use keywords or number values ‚Äã‚Äãto instruct CSS on how bold a piece of text should be.

As developers, when designing a website, sometimes we want to draw attention to something. There are several ways we can do this. The easiest way is to increase the character weight of the text you want to highlight. This refers to making the text appear bold.

In this article, we’ll talk about the weight font, how to define it, and provide an illustration of the different possible values ‚Äã‚Äãof the property.

CSS Fat: A Guide

The font-weight property sets are how bold text should appear on the screen. You can use words - keys or numeric value to say CSS how bold a text should appear together

The syntax of CSS font-weight property is as follows:.

The weightOfFont value is the weight of the font you want to use for the element to which the style is applied.

The font-weight property accepts a few different values, depending on the font weight you want to set for a particular element.

CSS scale bold

Think of your font family as having a scale of 100 to 1000 scale bold. The higher the number, the bolder the character

Do some research on your favorite font family. This will allow you to be sure that the result will be as light or bold as you want on your site. Some font families do not use full scale.

The following code shows how light and how dark a popular called font ’Arial’ can achieve in using numbers and words - key:

and bolder more clear: what do they actually mean?

The main thing that needs to be pointed out here is that lighter and bolder is not a literal translation. Bolder not meaning "bolder than bold"

Bolder and more clear in fact means, in this context, how bold they are or how clear they are in relation to their parent element. So when we do a more daring child element, they will be darker relative font than their parents - the same goes for the lighter.

The relative character weights are as follows:

td>
Relative character weight actual font weight
thin 100
normal 400
bold 700
heavy 900

So, we could use the weight font: bold property to set the weight font of a paragraph of text to 800. When a child a bolder relative weight of the parent, use the table above to understand how fat he is.

For example, let’s say I have a child who has a parent with a font-weight of 400. If I put the font-weight value on the bolder child, the weight font on the child becomes 700.

This is because the child is a darker font compared to -weight than the parent. If the parent is already as dark or as light as possible, these property values ‚Äã‚Äãdo nothing.

example CSS font weight

Let ’s say you’ re designing a website for The Seattle stamp club, a local stamp company. The stamp company we asked to do Who are - we en- head appear in bold on the A About their website. This will draw the visitor’s attention to the title.

The stamp club has asked us to add a block of text to their website with the history of the club. This block of text should appear at normal font weight. Some phrases that the club wants to draw the viewer’s attention to should appear in bold

We could use the following code to create this block of text, with some encouraged phrases:

Click style image of the button in the code editor above to see the output of our HTML / CSS code.

in our HTML file, we have defined a text enclosed paragraph of

; . tag we have also included some sentences in . tags, which we encourage in our CSS

Next, in our CSS file, we defined a style rule that defines the weight of the font style of each tag bolder . This means that text enclosed in a tag appears bolder than the parent element.

When we run our code, our paragraph appears with normal font width and sentences enclosed in tags are displayed in bold. In this example, the expressions welcomes everyone and 250 members are enclosed in tags.

Bold css text with variable characters

There are some new fonts made available to us by the latest level of CSS fonts. They are called variable characters and can take a number between 1 and 1000 as the character width. Browser support was not fully implemented until May 2020, so they’re quite new. If you want more information on variable fonts , check out this rel = "noopener"> l ’ introduction.

Conclusion

In this tutorial, we have discussed The CSS font -weight property and how it affects the boldness of our fonts.

We’ve learned that bolder and lighter weight means a relative font weight away from the parent element’s font weight . Basically, we looked at a range of character weights to see what the text looks like. You’ll be a pro font weighting in no time

To learn more about coding in CSS, read our guide on how to learn CSS . You can read our guide on how to make bold text in HTML if you want to make your text bold without relying on CSS.

Javascript Gets The Child Element By Tag 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).

Javascript Gets The Child Element By Tag 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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