67. 根据参数从右侧移除字符
string = "This is a sentence with "
# Remove trailing spaces from the right
print(string.rstrip()) # "This is a sentence with"
string = "this here is a sentence…..,,,,aaaaasd"
print(string.rstrip(“.,dsa”)) # "this here is a sentence"
string = "ffffffffFirst"
print(string.lstrip(“f”)) # First
68. 检查一个字符串是否代表一个数字
string = "seven"
print(string.isdigit()) # Falsestring = "1337"
print(string.isdigit()) # Truestring = "5a"
print(string.isdigit()) # False, because it contains the character 'a'string = "2**5"
print(string.isdigit()) # False
69. 检查一个字符串是否代表一个中文数字
# 42673 in Arabic numerals
string = "四二六七三"
print(string.isdigit()) # False
print(string.isnumeric()) # True
70. 检查一个字符串是否所有的单词都以大写字母开头
string = "This is a sentence"
print(string.istitle()) # False
string = "10 Python Tips"
print(string.istitle()) # True
string = "How to Print A String in Python"
# False, because of the first characters being lowercase in "to" and "in"
print(string.istitle())
string = "PYTHON"
print(string.istitle()) # False. It's titlelized version is "Python"
71. 我们也可以在元组中使用负索引
numbers = (1, 2, 3, 4)
print(numbers[-1]) # 4
print(numbers[-4]) # 1
72. 在元组中嵌套列表和元组
mixed_tuple = (("a"*10, 3, 4), ['first', 'second', 'third'])
print(mixed_tuple[1]) # ['first', 'second', 'third']
print(mixed_tuple[0]) # ('aaaaaaaaaa', 3, 4)
73. 快速计算满足条件的元素在列表中出现的次数
names = ["Besim", "Albert", "Besim", "Fisnik", "Meriton"]
print(names.count("Besim")) # 2
74. 使用slice()可以方便的得到最近的元素
my_list = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
slicing = slice(-4, None)
# Getting the last 3 elements from the list
print(my_list[slicing]) # [4, 5, 6]
# Getting only the third element starting from the right
print(my_list[-3]) # 4
你也可以使用*slice()*来完成其他常见的切片任务，比如:
string = "Data Science"
# start = 1, stop = None (don't stop anywhere), step = 1
# contains 1, 3 and 5 indices
slice_object = slice(5, None)
print(string[slice_object]) # Science
75. 计算元素在元组中出现的次数
my_tuple = ('a', 1, 'f', 'a', 5, 'a')
print(my_tuple.count('a')) # 3
76. 获取元组中元素的索引
my_tuple = ('a', 1, 'f', 'a', 5, 'a')
print(my_tuple.index('f')) # 2
77. 通过跳转获取子元组
my_tuple = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
print(my_tuple[::3]) # (1, 4, 7, 10)
78. 从索引开始获取子元组
my_tuple = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
print(my_tuple[3:]) # (4, 5, 6, 7, 8, 9, 10)
79. 从列表、集合或字典中删除所有元素
my_list = [1, 2, 3, 4]
my_list.clear()
print(my_list) # []
my_set = {1, 2, 3}
my_set.clear()
print(my_set) # set()
my_dict = {"a": 1, "b": 2}
my_dict.clear()
print(my_dict) # {}
80. 合并2个集合
一种方法是使用方法union()，它将作为合并的结果返回一个新的集合:
first_set = {4, 5, 6}
second_set = {1, 2, 3}
print(first_set.union(second_set)) # {1, 2, 3, 4, 5, 6}
另一个是方法update，它将第二个集合的元素插入到第一个集合中:
first_set = {4, 5, 6}
second_set = {1, 2, 3}
first_set.update(second_set)
print(first_set) # {1, 2, 3, 4, 5, 6}
81. 打印函数内的条件语句
def is_positive(number):
print("Positive" if number > 0 else "Negative") # Positive
is_positive(-3)
82. 一个if语句中包含多个条件
math_points = 51
biology_points = 78
physics_points = 56
history_points = 72
my_conditions = [math_points > 50, biology_points > 50,
physics_points > 50, history_points > 50]
if all(my_conditions):
print("Congratulations! You have passed all of the exams.")
else:
print("I am sorry, but it seems that you have to repeat at least one exam.")
# Congratulations! You have passed all of the exams.
83. 在一个if语句中至少满足一个条件
math_points = 51
biology_points = 78
physics_points = 56
history_points = 72
my_conditions = [math_points > 50, biology_points > 50,
physics_points > 50, history_points > 50]
if any(my_conditions):
print("Congratulations! You have passed all of the exams.")
else:
print("I am sorry, but it seems that you have to repeat at least one exam.")
# Congratulations! You have passed all of the exams.
84. 任何非空字符串都被计算为True
print(bool("Non empty")) # True
print(bool("")) # False
85. 任何非空列表、元组或字典都被求值为True
print(bool([])) # False
print(bool(set([]))) # False
print(bool({})) # False
print(bool({"a": 1})) # True
86. 其他计算为False的值是None、“False”和数字0
print(bool(False)) # False
print(bool(None)) # False
print(bool(0)) # False
87. 你不能仅仅通过在函数中提及全局变量来改变它的值
string = "string"
def do_nothing():
string = "inside a method"
do_nothing()
print(string) # string
string = "string"
def do_nothing():
global string
string = "inside a method"
do_nothing()
print(string) # inside a method
88. 使用“collections”中的Counter计算字符串或列表中的元素数量
from collections import Counter
result = Counter("Banana")
print(result) # Counter({'a': 3, 'n': 2, 'B': 1})
result = Counter([1, 2, 1, 3, 1, 4, 1, 5, 1, 6])
print(result) # Counter({1: 5, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1})
89. 使用Counter检查是否2个字符串包含相同的字符
from collections import Counter
def check_if_anagram(first_string, second_string):
first_string = first_string.lower()
second_string = second_string.lower()
return Counter(first_string) == Counter(second_string)
print(check_if_anagram('testinG', 'Testing')) # True
print(check_if_anagram('Here', 'Rehe')) # True
print(check_if_anagram('Know', 'Now')) # False
你也可以使用*sorted()*检查两个字符串是否具有相同的字符：
def check_if_anagram(first_word, second_word):
first_word = first_word.lower()
second_word = second_word.lower()
return sorted(first_word) == sorted(second_word)
print(check_if_anagram("testinG", "Testing")) # True
print(check_if_anagram("Here", "Rehe")) # True
print(check_if_anagram("Know", "Now")) # False
90. 使用" itertools "中的" Count "计算元素的数量
from itertools import count
my_vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
current_counter = count()
string = "This is just a sentence."
for i in string:
if i in my_vowels:
print(f"Current vowel: {i}")
print(f"Number of vowels found so far: {next(current_counter)}")
Current vowel: i
Number of vowels found so far: 0
Current vowel: i
Number of vowels found so far: 1
Current vowel: u
Number of vowels found so far: 2
Current vowel: a
Number of vowels found so far: 3
Current vowel: e
Number of vowels found so far: 4
Current vowel: e
Number of vowels found so far: 5
Current vowel: e
Number of vowels found so far: 6
91. 根据字符串或列表的频率对元素进行排序
来自collections模块的Counter默认情况下不会根据元素的频率来排序。
from collections import Counter
result = Counter([1, 2, 3, 2, 2, 2, 2])
print(result) # Counter({2: 5, 1: 1, 3: 1})
print(result.most_common()) # [(2, 5), (1, 1), (3, 1)]
92. 在一行中找到列表中出现频次最高的元素
my_list = ['1', 1, 0, 'a', 'b', 2, 'a', 'c', 'a']
print(max(set(my_list), key=my_list.count)) # a
93. copy()和deepcopy()的区别
浅拷贝构造一个新的复合对象，然后(在可能的范围内)在其中插入对原始对象的引用。深拷贝构造一个新的复合对象，然后递归地将在原始对象中找到的对象的副本插入其中。
浅拷贝意味着构造一个新的集合对象，然后用对原始集合中的子对象的引用填充它。从本质上说，浅拷贝的深度只有一层。拷贝过程不会递归，因此不会创建子对象本身的副本。深拷贝使拷贝过程递归。这意味着首先构造一个新的集合对象，然后用在原始集合对象中找到的子对象的副本递归地填充它。以这种方式拷贝对象将遍历整个对象树，以创建原始对象及其所有子对象的完全独立的克隆。
first_list = [[1, 2, 3], ['a', 'b', 'c']]
second_list = first_list.copy()
first_list[0][2] = 831
print(first_list) # [[1, 2, 831], ['a', 'b', 'c']]
print(second_list) # [[1, 2, 831], ['a', 'b', 'c']]
import copy
first_list = [[1, 2, 3], ['a', 'b', 'c']]
second_list = copy.deepcopy(first_list)
first_list[0][2] = 831
print(first_list) # [[1, 2, 831], ['a', 'b', 'c']]
print(second_list) # [[1, 2, 3], ['a', 'b', 'c']]
94. 当试图访问字典中不存在的键时，可以避免抛出错误
如果你使用一个普通的字典，并试图访问一个不存在的键，那么你将得到一个错误:
my_dictonary = {"name": "Name", "surname": "Surname"}print(my_dictonary["age"])
我们可以使用defaultdict()：来避免这种错误
from collections import defaultdict
my_dictonary = defaultdict(str)
my_dictonary['name'] = "Name"
my_dictonary['surname'] = "Surname"
print(my_dictonary["age"])
95. 你可以构建自己的迭代器
class OddNumbers:
def __iter__(self):
self.a = 1
return self
def __next__(self):
x = self.a
self.a += 2
return x
odd_numbers_object = OddNumbers()
iterator = iter(odd_numbers_object)
print(next(iterator)) # 1
print(next(iterator)) # 3
print(next(iterator)) # 5
96. 可以用一行从列表中删除重复项
my_set = set([1, 2, 1, 2, 3, 4, 5])
print(list(my_set)) # [1, 2, 3, 4, 5]
97. 打印模块所在的位置
import torch
print(torch) # <module 'torch' from '/Users/...'
98. 可以使用" not in "来检查值是否不属于列表
odd_numbers = [1, 3, 5, 7, 9]
even_numbers = []
for i in range(9):
if i not in odd_numbers:
even_numbers.append(i)
print(even_numbers) # [0, 2, 4, 6, 8]
99. sort() 和 sorted()的差别
groceries = ['milk', 'bread', 'tea']
new_groceries = sorted(groceries)
# new_groceries = ['bread', 'milk', 'tea']
print(new_groceries)
# groceries = ['milk', 'bread', 'tea']
print(groceries)
groceries.sort()
# groceries = ['bread', 'milk', 'tea']
print(groceries)
100. 使用uuid模块生成唯一的id
import uuid
# Generate a UUID from a host ID, sequence number, and the current time
print(uuid.uuid1()) # 308490b6-afe4-11eb-95f7-0c4de9a0c5af
# Generate a random UUID
print(uuid.uuid4()) # 93bc700b-253e-4081-a358-24b60591076a
