Javascript Charat Method

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There is a String object method in JavaScript that allows us to determine which character is at a given index in a string. This article will tell you about this method and show you how to use it.

For this method, we turn to our primitive types in JavaScript. What are these guys? What does primitive also mean ?

Primitive types, or values, are elements that are not objects and do not have associated methods. In JavaScript, the primitives are string, number, boolean, bigint, symbol, and undefined. Today we are going to focus on the primitive chain value.

The interesting thing about the primitive string value is that it is usually interchangeable with the primitive String object. In most cases, JavaScript will automatically wrap a primitive String object wrapper around a primitive string and force the use of String methods. So in most cases, you don’t have to worry about changing a primitive string (with no methods available) to a primitive String object (with methods available).

Syntax

The syntax of the String charAt () method is quite simple. Let’s start with the channel we want to look at. Since this is a String object, we use dot notation to access the method and bind it at the end. To invoke it, we add parentheses at the end with the index we want to find.

The charAt () method returns the character at this index in the string. Remember that strings are zero-based, so indices start at 0 and start from there in sequential order. Here’s a working implementation - use inputs to add a string and index, then hit enter. Unless you requested an out-of-range index, it should return a single character in the requested index.


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Javascript Charat Method around: Questions

Removing white space around a saved image in matplotlib

2 answers

I need to take an image and save it after some process. The figure looks fine when I display it, but after saving the figure, I got some white space around the saved image. I have tried the "tight" option for savefig method, did not work either. The code:

  import matplotlib.image as mpimg
  import matplotlib.pyplot as plt

  fig = plt.figure(1)
  img = mpimg.imread(path)
  plt.imshow(img)
  ax=fig.add_subplot(1,1,1)

  extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
  plt.savefig("1.png", bbox_inches=extent)

  plt.axis("off") 
  plt.show()

I am trying to draw a basic graph by using NetworkX on a figure and save it. I realized that without a graph it works, but when added a graph I get white space around the saved image;

import matplotlib.image as mpimg
import matplotlib.pyplot as plt
import networkx as nx

G = nx.Graph()
G.add_node(1)
G.add_node(2)
G.add_node(3)
G.add_edge(1,3)
G.add_edge(1,2)
pos = {1:[100,120], 2:[200,300], 3:[50,75]}

fig = plt.figure(1)
img = mpimg.imread("image.jpg")
plt.imshow(img)
ax=fig.add_subplot(1,1,1)

nx.draw(G, pos=pos)

extent = ax.get_window_extent().transformed(fig.dpi_scale_trans.inverted())
plt.savefig("1.png", bbox_inches = extent)

plt.axis("off") 
plt.show()
228

Answer #1

You can remove the white space padding by setting bbox_inches="tight" in savefig:

plt.savefig("test.png",bbox_inches="tight")

You"ll have to put the argument to bbox_inches as a string, perhaps this is why it didn"t work earlier for you.


Possible duplicates:

Matplotlib plots: removing axis, legends and white spaces

How to set the margins for a matplotlib figure?

Reduce left and right margins in matplotlib plot

228

Answer #2

I cannot claim I know exactly why or how my “solution” works, but this is what I had to do when I wanted to plot the outline of a couple of aerofoil sections — without white margins — to a PDF file. (Note that I used matplotlib inside an IPython notebook, with the -pylab flag.)

plt.gca().set_axis_off()
plt.subplots_adjust(top = 1, bottom = 0, right = 1, left = 0, 
            hspace = 0, wspace = 0)
plt.margins(0,0)
plt.gca().xaxis.set_major_locator(plt.NullLocator())
plt.gca().yaxis.set_major_locator(plt.NullLocator())
plt.savefig("filename.pdf", bbox_inches = "tight",
    pad_inches = 0)

I have tried to deactivate different parts of this, but this always lead to a white margin somewhere. You may even have modify this to keep fat lines near the limits of the figure from being shaved by the lack of margins.

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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