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1:将两个列表兼并成一个字典
假定咱们在 Python 中有两个列表,咱们期望将它们兼并为字典形式,其间一个列表的项作为字典的键,另一个作为值。这是在用 Python 编写代码时经常遇到的一个十分常见的问题。
可是为了解决这个问题,咱们需求考虑几个约束,比如两个列表的巨细,两个列表中元素的类型,以及其间是否有重复的元素,尤其是咱们将运用的元素作为 key 时。咱们能够经过运用 zip 等内置函数来解决这些问题
keys_list = ['A', 'B', 'C']
values_list = ['blue', 'red', 'bold']
#There are 3 ways to convert these two lists into a dictionary
#1- Using Python's zip, dict functionz
dict_method_1 = dict(zip(keys_list, values_list))
#2- Using the zip function with dictionary comprehensions
dict_method_2 = {key:value for key, value in zip(keys_list, values_list)}
#3- Using the zip function with a loop
items_tuples = zip(keys_list, values_list)
dict_method_3 = {}
for key, value in items_tuples:
if key in dict_method_3:
pass # To avoid repeating keys.
else:
dict_method_3[key] = value
2:将两个或多个列表兼并为一个包括列表的列表
另一个常见的使命是当咱们有两个或更多列表时,咱们期望将它们悉数收集到一个大列表中,其间较小列表的所有第一项构成较大列表中的第一个列表
例如,假如咱们有 4 个列表 [1,2,3], [‘a’,’b’,’c’], [‘h’,’e’,’y’] 和 [4,5, 6],咱们想为这四个列表创建一个新列表;它将是 [[1,’a’,’h’,4], [2,’b’,’e’,5], [3,’c’,’y’,6]]
def merge(*args, missing_val = None):
#missing_val will be used when one of the smaller lists is shorter tham the others.
#Get the maximum length within the smaller lists.
max_length = max([len(lst) for lst in args])
outList = []
for i in range(max_length):
result.append([args[k][i] if i < len(args[k]) else missing_val for k in range(len(args))])
return outList
3:对字典列表进行排序
这一组日常列表使命是排序使命,根据列表中包括的元素的数据类型,咱们将采用稍微不同的方式对它们进行排序。
dicts_lists = [
{
"Name": "James",
"Age": 20,
},
{
"Name": "May",
"Age": 14,
},
{
"Name": "Katy",
"Age": 23,
}
]
#There are different ways to sort that list
#1- Using the sort/ sorted function based on the age
dicts_lists.sort(key=lambda item: item.get("Age"))
#2- Using itemgetter module based on name
from operator import itemgetter
f = itemgetter('Name')
dicts_lists.sort(key=f)
4:对字符串列表进行排序
咱们经常面临包括字符串的列表,咱们需求按字母次序、长度或咱们想要或咱们的应用程序需求的任何其他要素对这些列表进行排序
my_list = ["blue", "red", "green"]
#1- Using sort or srted directly or with specifc keys
my_list.sort() #sorts alphabetically or in an ascending order for numeric data
my_list = sorted(my_list, key=len) #sorts the list based on the length of the strings from shortest to longest.
# You can use reverse=True to flip the order
#2- Using locale and functools
import locale
from functools import cmp_to_key
my_list = sorted(my_list, key=cmp_to_key(locale.strcoll))
5:根据另一个列表对列表进行排序
有时,咱们或许需求运用一个列表来对另一个列表进行排序,因而,咱们将有一个数字列表(索引)和一个咱们想运用这些索引进行排序的列表
a = ['blue', 'green', 'orange', 'purple', 'yellow']
b = [3, 2, 5, 4, 1]
#Use list comprehensions to sort these lists
sortedList = [val for (_, val) in sorted(zip(b, a), key=lambda x: \
x[0])]
6:将列表映射到字典
列表代码片段的最后一个使命,假如给定一个列表并将其映射到字典中,也便是说,咱们想将咱们的列表转换为带有数字键的字典
mylist = ['blue', 'orange', 'green']
#Map the list into a dict using the map, zip and dict functions
mapped_dict = dict(zip(itr, map(fn, itr)))
Dictionary Snippets
现在处理的数据类型是字典
7:兼并两个或多个字典
假定咱们有两个或多个字典,而且咱们期望将它们悉数兼并为一个具有唯一键的字典
from collections import defaultdict
#merge two or more dicts using the collections module
def merge_dicts(*dicts):
mdict = defaultdict(list)
for dict in dicts:
for key in dict:
res[key].append(d[key])
return dict(mdict)
8:回转字典
一个十分常见的字典使命是假如咱们有一个字典而且想要翻转它的键和值,键将成为值,而值将成为键
当咱们这样做时,咱们需求保证没有重复的键。值能够重复,但键不能,并保证所有新键都是能够 hashable 的
my_dict = {
"brand": "Ford",
"model": "Mustang",
"year": 1964
}
#Invert the dictionary based on its content
#1- If we know all values are unique.
my_inverted_dict = dict(map(reversed, my_dict.items()))
#2- If non-unique values exist
from collections import defaultdict
my_inverted_dict = defaultdict(list)
{my_inverted_dict[v].append(k) for k, v in my_dict.items()}
#3- If any of the values are not hashable
my_dict = {value: key for key in my_inverted_dict for value in my_inverted_dict[key]}
String Snippets
接下来是字符串的处理
9:运用 f 字符串
格式化字符串或许是咱们简直每天都需求完成的一项使命,在 Python 中有多种方法能够格式化字符串,运用 f 字符串是比较好的挑选
#Formatting strings with f string.
str_val = 'books'
num_val = 15
print(f'{num_val} {str_val}') # 15 books
print(f'{num_val % 2 = }') # 1
print(f'{str_val!r}') # books
#Dealing with floats
price_val = 5.18362
print(f'{price_val:.2f}') # 5.18
#Formatting dates
from datetime import datetime;
date_val = datetime.utcnow()
print(f'{date_val=:%Y-%m-%d}') # date_val=2021-09-24
10:查看子串
一项十分常见的使命便是查看字符串是否在与字符串列表中
addresses = ["123 Elm Street", "531 Oak Street", "678 Maple Street"]
street = "Elm Street"
#The top 2 methods to check if street in any of the items in the addresses list
#1- Using the find method
for address in addresses:
if address.find(street) >= 0:
print(address)
#2- Using the "in" keyword
for address in addresses:
if street in address:
print(address)
11:以字节为单位获取字符串的巨细
有时,尤其是在构建内存要害应用程序时,咱们需求知道咱们的字符串运用了多少内存
str1 = "hello"
str2 = "你好"
def str_size(s):
return len(s.encode('utf-8'))
str_size(str1)
str_size(str2)
12:Input/ Output operations
最后咱们来看看输入输出方面的代码片段
№12:查看文件是否存在
在数据科学和许多其他应用程序中,咱们经常需求从文件中读取数据或向其间写入数据,但要做到这一点,咱们需求查看文件是否存在,因而,咱们需求保证代码不会因 IO 过错而停止
#Checking if a file exists in two ways
#1- Using the OS module
import os
exists = os.path.isfile('/path/to/file')
#2- Use the pathlib module for a better performance
from pathlib import Path
config = Path('/path/to/file')
if config.is_file():
pass
13:解析电子表格
另一种十分常见的文件交互是从电子表格中解析数据,咱们运用 CSV 模块来帮助咱们有效地执行该使命
import csv
csv_mapping_list = []
with open("/path/to/data.csv") as my_data:
csv_reader = csv.reader(my_data, delimiter=",")
line_count = 0
for line in csv_reader:
if line_count == 0:
header = line
else:
row_dict = {key: value for key, value in zip(header, line)}
csv_mapping_list.append(row_dict)
line_count += 1