你需要知道的20个常用的Python技巧
小白学视觉
共 9258字,需浏览 19分钟
·
2024-05-02 10:05
重磅干货,第一时间送达
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重磅干货,第一时间送达
来源:https://medium.com
使用Python切片反转字符串:
# Reversing a string using slicing
my_string = "ABCDE"
reversed_string = my_string[::-1]
print(reversed_string)
# Output
# EDCBA
使用title函数方法:
my_string = "my name is chaitanya baweja"
# using the title() function of string class
new_string = my_string.title()
print(new_string)
# Output
# My Name Is Chaitanya Baweja
使用集合的概念查找字符串的唯一元素:
my_string = "aavvccccddddeee"
# converting the string to a set
temp_set = set(my_string)
# stitching set into a string using join
new_string = ''.join(temp_set)
print(new_string)
# output
# cdvae
你可以使用乘法符号(*)打印字符串或列表多次:
n = 3 # number of repetitions
my_string = "abcd"
my_list = [1,2,3]
print(my_string*n)
# abcdabcdabcd
print(my_list*n)
# [1,2,3,1,2,3,1,2,3]
# Multiplying each element in a list by 2
original_list = [1,2,3,4]
new_list = [2*x for x in original_list]
print(new_list)
# [2,4,6,8]
a = 1
b = 2
a, b = b, a
print(a) # 2
print(b) # 1
使用.split()函数:
string_1 = "My name is Chaitanya Baweja"
string_2 = "sample/ string 2"
# default separator ' '
print(string_1.split())
# ['My', 'name', 'is', 'Chaitanya', 'Baweja']
# defining separator as '/'
print(string_2.split('/'))
# ['sample', ' string 2']
list_of_strings = ['My', 'name', 'is', 'Chaitanya', 'Baweja']
# Using join with the comma separator
print(','.join(list_of_strings))
# Output
# My,name,is,Chaitanya,Baweja
my_string = "abcba"
if my_string == my_string[::-1]:
print("palindrome")
else:
print("not palindrome")
# Output
# palindrome
# finding frequency of each element in a list
from collections import Counter
my_list = ['a','a','b','b','b','c','d','d','d','d','d']
count = Counter(my_list) # defining a counter object
print(count) # Of all elements
# Counter({'d': 5, 'b': 3, 'a': 2, 'c': 1})
print(count['b']) # of individual element
# 3
print(count.most_common(1)) # most frequent element
# [('d', 5)]
Anagrams的含义为两个单词中,每个英文单词(不含大小写)出现的次数相同,使用Counter类判断两个字符串是否为Anagrams。
from collections import Counter
str_1, str_2, str_3 = "acbde", "abced", "abcda"
cnt_1, cnt_2, cnt_3 = Counter(str_1), Counter(str_2), Counter(str_3)
if cnt_1 == cnt_2:
print('1 and 2 anagram')
if cnt_1 == cnt_3:
print('1 and 3 anagram')
# output
# 1 and 2 anagram
except获取异常处理:
a, b = 1,0
try:
print(a/b)
# exception raised when b is 0
except ZeroDivisionError:
print("division by zero")
else:
print("no exceptions raised")
finally:
print("Run this always")
# output
# division by zero
# Run this always
my_list = ['a', 'b', 'c', 'd', 'e']
for index, value in enumerate(my_list):
print('{0}: {1}'.format(index, value))
# 0: a
# 1: b
# 2: c
# 3: d
# 4: e
import sys
num = 21
print(sys.getsizeof(num))
# In Python 2, 24
# In Python 3, 28
dict_1 = {'apple': 9, 'banana': 6}
dict_2 = {'banana': 4, 'orange': 8}
combined_dict = {**dict_1, **dict_2}
print(combined_dict)
# Output
# {'apple': 9, 'banana': 4, 'orange': 8}
使用time类计算运行一段代码所花费的时间:
import time
start_time = time.time()
# Code to check follows
for i in range(10**5):
a, b = 1,2
c = a+ b
# Code to check ends
end_time = time.time()
time_taken_in_micro = (end_time- start_time)*(10**6)
print(time_taken_in_micro)
# output
# 18770.217895507812
from iteration_utilities import deepflatten
# if you only have one depth nested_list, use this
def flatten(l):
return [item for sublist in l for item in sublist]
l = [[1,2,3],[3]]
print(flatten(l))
# [1, 2, 3, 3]
# if you don't know how deep the list is nested
l = [[1,2,3],[4,[5],[6,7]],[8,[9,[10]]]]
print(list(deepflatten(l, depth=3)))
# [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
import random
my_list = ['a', 'b', 'c', 'd', 'e']
num_samples = 2
samples = random.sample(my_list,num_samples)
print(samples)
# [ 'a', 'e'] this will have any 2 random values
将整数转化成数字列表:
num = 123456
# using map
list_of_digits = list(map(int, str(num)))
print(list_of_digits)
# [1, 2, 3, 4, 5, 6]
# using list comprehension
list_of_digits = [int(x) for x in str(num)]
print(list_of_digits)
# [1, 2, 3, 4, 5, 6]
检查列表中每个元素是否为唯一的:
def unique(l):
if len(l)==len(set(l)):
print("All elements are unique")
else:
print("List has duplicates")
unique([1,2,3,4])
# All elements are unique
unique([1,1,2,3])
# List has duplicates
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