Javascript Replaces The Object In The Array

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There are three ways to replace an element in a Python list. You can use list indexing or a loop to replace an item. If you want to create a new list based on an existing list and make a change, you can use a list comprehension .

You can decide to modify a value in a list. Let’s say you are creating a menu for a restaurant. You may notice that you spelled one of the wrong menu items. To correct this error, you need to modify a You can replace an item in a Python list by using an existing item loop in the list.

Python Replace element in list

, index lists o comprehension lists . The first two methods modify an existing list into a list comprehension creates a new list with the specified modifications

Either of the Summarize each method:.

  • Indexing lists: we use the index number of an element in the list to modify the value associated with this element. The equal sign is used to modify a value in a list
  • List Include:. List syntax Understand creates a new list of an existing company . You can specify conditions in the understanding of the list to determine the values ‚Äã‚Äãin the new list
  • for the loop. The loop iterates over the elements of a list. Indexing is used to make the actual change to the list. We use the enumerate () method to create two lists of index numbers and values ‚Äã‚Äãthat we can iterate over.

In this guide, we take a look at each of these methods. We will refer to an example of each to help you get started

Python Replace item in list. By using the list of indexing

The best way to replace an item in a list is uses Python syntax indexing . Indexing allows you to choose an item or a series of items from a list. With the assignment operator, you can change a value at a certain position in a list.

We are building a store price information program in a clothing store. the price of the first item on our list should be increased by $ 5. To do this we use the ’indexing of lists

Either start by creating a list containing the prices of our products:.

We use indexing to select and modify the first point in our list, 99.95. This value has the index position zero. This is because the lists are indexed from zero:

Our code selects the element at position zero and sets its value to 104.95. This represents an increase of $ 5 on the old value. Our code returns our list of items with the first price change:

We can modify our list by adding five to the current value of prices [0]:

price [0] corresponds to first element of our list (the one at position index)

our code returns a list with the same values ‚Äã‚Äãas our first example:.

Python Replace element in list: using a list including

Introducing the Python list may be the most Python way to find and replace an item in a list. This method is useful if you want to create a new list based on the values ‚Äã‚Äãof an existing one.

Using a List Understanding allows you to browse through items in an existing list and create a new list based on a particular criteria. You can generate a new list that contains items that start with "C" from an existing list, for example.

Here we create a program that calculates a discount 10% on all products in a store selling retail. of clothing worth more than $ 50 We use our previous price list of products:

Then, we define a list comprising to replace the elements of our list:

This list including iterates through the "price" list and finds values ‚Äã‚Äãworth more than $ 50. A 10% discount is applied to these items. Let’s wrap around the updated values ‚Äã‚Äãto two decimal places using the round () method

Our code returns our list of new prices.

A reduction was successfully applied to each element of 10%

Python replace the element in the list. using a For loop

You can replace elements in a list using a python by loop . To do this, we need Python enumerate () function . This function returns two lists: index numbers in a list and values ‚Äã‚Äãin a list. Let iterate over these two lists with a single loop

In this example, we’ll use the same price list in our code:.

we then define a loop that iterates over this list with the enumerate function () :

The value "index" stores the position of the index d ’an element. "Item" is the value linked to this index position. The comma separates the two list values ‚Äã‚Äãreturned by the enumerate () method

Retrieving two or more values ‚Äã‚Äãof a method or another value is called decompression. We have "unzipped" two lists of the enumerate () method.

We use the same formula above to calculate a value of over $ 50 off 10% items. Let’s run our code and see what happens:

Our code correctly changes the items in our "price" list based on our discount.

Conclusion

You can replace items in a Python list using the understanding of list indexing, or looping.

If you want to replace a value in a list, the indexing syntax is more appropriate. To replace multiple items in a list that meet a criterion, using an understanding of the list is a good solution. While for the curls are functional, they are less Pythonic than the comprehension of the list.

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Javascript Replaces The Object In The Array 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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