Vue3+Vite+TS implements secondary encapsulation of element-plus business components sfasga

1. Structure string

You will often need to print strings. If you have many variables, avoid the following:

name = "Raymond"
age = 22
born_in = "Oakland, CA"
string = "Hello my name is " + name + "and I'm " + str(age) + " years old. I was born in " + born_in + "."
print(string)

How messy does this look? You can use a nice and simple alternative to .format.

as follows:

name = "Raymond"
age = 22
born_in = "Oakland, CA"
string = "Hello my name is {0} and I'm {1} years old. I was born in {2}.".format(name, age, born_in)
print(string)

2. Return tuple

Python allows you to return multiple elements from a function, which makes life much easier. But when unpacking tuples, I make a common mistake like this:

def binary(): return 0, 1
result = binary()
zero = result[0]
one = result[1]

This is not necessary, you can just do this:

def binary(): return 0, 1
zero, one = binary()

If you need all elements to be returned, use an underscore _

zero, _ = binary()

It’s so efficient!

3. Access Dict dictionary

You will also often write key , value to dicts .

If you try to access a key that doesn't exist in dict , you might be tempted to do this to avoid KeyError :

countr = {}
bag = [2, 3, 1, 2, 5, 6, 7, 9, 2, 7]
for i in bag:
if i in countr:
countr[i] += 1 else:
countr[i] = 1
for i in range(10):
if i in countr:
print("Count of {}: {}".format(i, countr[i]))
else:
print("Count of {}: {}".format(i, 0))

However, using get() is a better approach.

countr = {}
bag = [2, 3, 1, 2, 5, 6, 7, 9, 2, 7]
for i in bag:
countr[i] = countr.get(i, 0) + 1
for i in range(10):
print("Count of {}: {}".format(i, countr.get(i, 0)))

Of course you can also use setdefault instead.

There is also a simpler but more expensive method:

bag = [2, 3, 1, 2, 5, 6, 7, 9, 2, 7]

{2: 3, 3: 1, 1: 1, 5: 1, 6: 1, 7: 2, 9: 1}:

countr = dict([(num, bag.count(num)) for num in bag])
for i in range(10):
print("Count of {}: {}".format(i, countr.get(i, 0)))

We can also use dict comprehensions.

countr = {num: bag.count(num) for num in bag}

Both of these methods are expensive because they traverse the list every time count is called.

4. Use the library

Existing libraries can be imported to do exactly what you want.

Let's go back to the previous example and create a function to count the number of times a number appears in a list. Well, there used to be a library that could do this.

from collections import Counter
bag = [2, 3, 1, 2, 5, 6, 7, 9, 2, 7]
countr = Counter(bag)for i in range(10):
print("Count of {}: {}".format(i, countr[i]))

Some reasons to use libraries:

• The code is correct and tested.
• Their algorithms may be optimal, so they run faster.
• Abstract: They are clear and well documented, and you can focus on what is yet to be done.
  

In the end, it was all there before, you don't have to reinvent the wheel.

5. Slicing/stepping through lists

We can specify start point and the stop point, like this list[start:stop:step] . We take the first 5 elements of the list:

bag = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for elem in bag[:5]:
print(elem)

This is slicing. We specify stop point as 5, and then 5 elements will be taken from the list before stopping.

What if the last 5 elements?

bag = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for elem in bag[-5:]:
print(elem)

Don’t you understand? -5 means take 5 elements from the end of the list.

If you want to perform a distance operation on the elements in a list, you might do this:

bag = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for index, elem in enumerate(bag):
if index % 2 == 0:
print(elem)

But you should do it like this:

bag = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
for elem in bag[::2]:
print(elem)

6. Use ranges

bag = list(range(0,10,2))
print(bag)

This is the stepping in the list. list[::2] means traversing the list and taking out an element in two steps.

You can do cool reverse operations on a list with list[::-1] .

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