17条提高工作效率的Python技巧分享

# bad practice码
n1 = input("enter a number : ")
n2 = input("enter a number : ")
n2 = input("enter a number : ")
print(n1, n2, n3)

# good practice
n1, n2, n3 = input("enter a number : ").split()
print(n1, n2, n3)

size = "lg"
color = "blue"
price = 50
# bad practice
if size == "lg" and color == "blue" and price < 100:
    print("Yes, I want to but the product.")

# good practice
conditions = [
    size == "lg",
    color == "blue",
    price < 100,
]
if all(conditions):
    print("Yes, I want to but the product.")

# bad practice
size = "lg"
color = "blue"
price = 50
if size == "lg" or color == "blue" or price < 100:
    print("Yes, I want to but the product.")

# good practice
conditions = [
    size == "lg",
    color == "blue",
    price < 100,
]
if any(conditions):
    print("Yes, I want to but the product.")

print('odd' if int(input('Enter a number: '))%2 else 'even')

v1 = 100
v2 = 200
# bad practice
temp = v1
v1 = v2
v2 = temp

v1 = 100
v2 = 200
# good practice
v1, v2 = v2, v1

print("John Deo"[::-1])

v1 = "madam" # is a palindrome string
v2 = "master" # is not a palindrome string
print(v1.find(v1[::-1]) == 0) # True
print(v1.find(v2[::-1]) == 0) # False

name = "ali"
age = 22
# bad practices
if name:
    print(name)
if name and age > 18:
    print("user is verified")

# a better approach
print(name if name else "")
""" here you have to define the else condition too"""
# good practice 
name and print(name)
age > 18 and name and print("user is verified")

lst = [1, 2, 3, 4, 3, 4, 4, 5, 6, 3, 1, 6, 7, 9, 4, 0]
print(lst)
unique_lst = list(set(lst))
print(unique_lst)

lst = [1, 2, 3, 4, 3, 4, 4, 5, 6, 3, 1, 6, 7, 9, 4, 0]
most_repeated_item = max(lst, key=lst.count)
print(most_repeated_item)

numbers = [1,2,3,4,5,6,7]
evens = [x for x in numbers if x % 2 is 0]
odds = [y for y in numbers if y not in evens]
cities = ['London', 'Dublin', 'Oslo']
def visit(city):
    print("Welcome to "+city)
for city in cities:
    visit(city)

def sum_of_squares(n1, n2)
    return n1**2 + n2**2
print(sum_of_squares(2,3))
# output: 13
"""
what ever if you want to pass, multiple args to the function 
as n number of args. so let's make it dynamic.
""" 
def sum_of_squares(*args):
    return sum([item**2 for item in args])
# now you can pass as many parameters as you want
print(sum_of_squares(2, 3, 4))
print(sum_of_squares(2, 3, 4, 5, 6))

lst = ["blue", "lightblue", "pink", "orange", "red"]
for idx, item in enumerate(lst):
     print(idx, item)

names = ["john", "sara", "jim", "rock"]
print(", ".join(names))

d1 = {"v1": 22, "v2": 33}
d2 = {"v2": 44, "v3": 55}
d3 = {**d1, **d2}
print(d3)

keys = ['a', 'b', 'c']
vals = [1, 2, 3]
zipped = dict(zip(keys, vals))

d = {
    "v1": 80,
    "v2": 20,
    "v3": 40,
    "v4": 20,
    "v5": 10,
}
sorted_d = dict(sorted(d.items(), key=lambda item: item[1]))
print(sorted_d)

from operator import itemgetter
sorted_d = dict(sorted(d.items(), key=itemgetter(1)))

sorted_d = dict(sorted(d.items(), key=itemgetter(1), reverse=True))

from pprint import pprint
data = {
    "name": "john deo",
    "age": "22",
    "address": {"contry": "canada", "state": "an state of canada :)", "address": "street st.34 north 12"},
    "attr": {"verified": True, "emialaddress": True},
}
print(data)
pprint(data)

{'name': 'john deo', 'age': '22', 'address': {'contry': 'canada', 'state': 'an state of canada :)', 'address': 'street st.34 north 12'}, 'attr': {'verified': True, 'emialaddress': True}}
{'address': {'address': 'street st.34 north 12',
             'contry': 'canada',
             'state': 'an state of canada :)'},
 'age': '22',
 'attr': {'emialaddress': True, 'verified': True},
 'name': 'john deo'}