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以下文章来源于快学Python ,作者快快
zip 这样的内置函数来克服这个问题。keys_list = ['A', 'B', 'C']
values_list = ['blue', 'red', 'bold']
# 有 3 种方法可以将这两个列表转换为字典
# 1.使用Python zip、dict函数
dict_method_1 = dict(zip(keys_list, values_list))
# 2. 使用带有字典推导式的 zip 函数
dict_method_2 = {key:value for key, value in zip(keys_list, values_list)}
# 3.循环使用zip函数
items_tuples = zip(keys_list, values_list)
dict_method_3 = {}
for key, value in items_tuples:
if key in dict_method_3:
pass
else:
dict_method_3[key] = value
print(dict_method_1)
print(dict_method_2)
print(dict_method_3)

[1,2,3]、['a','b','c']、['h','e','y'], 和[4,5,6],我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):
max_length = max([len(lst) for lst in args])
outList = []
for i in range(max_length):
outList.append([args[k][i] if i < len(args[k]) else missing_val for k in range(len(args))])
return outList
merge([1,2,3],['a','b','c'],['h','e','y'],[4,5,6])

dicts_lists = [
{
"Name": "James",
"Age": 20,
},
{
"Name": "May",
"Age": 14,
},
{
"Name": "Katy",
"Age": 23,
}
]
# 方法一
dicts_lists.sort(key=lambda item: item.get("Age"))
# 方法二
from operator import itemgetter
f = itemgetter('Name')
dicts_lists.sort(key=f)

my_list = ["blue", "red", "green"]
# 方法一
my_list.sort()
my_list = sorted(my_list, key=len)
# 方法二
import locale
from functools import cmp_to_key
my_list = sorted(my_list, key=cmp_to_key(locale.strcoll))

a = ['blue', 'green', 'orange', 'purple', 'yellow']
b = [3, 2, 5, 4, 1]
sortedList = [val for (_, val) in sorted(zip(b, a), key=lambda x: x[0])]
print(sortedList)

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)))
from collections import defaultdict
def merge_dicts(*dicts):
mdict = defaultdict(list)
for dict in dicts:
for key in dict:
res[key].append(d[key])
return dict(mdict)
my_dict = {
"brand": "Ford",
"model": "Mustang",
"year": 1964
}
# 方法一
my_inverted_dict_1 = dict(map(reversed, my_dict.items()))
# 方法二
from collections import defaultdict
my_inverted_dict_2 = defaultdict(list)
{my_inverted_dict_2[v].append(k) for k, v in my_dict.items()}
print(my_inverted_dict_1)
print(my_inverted_dict_2)

str_val = 'books'
num_val = 15
print(f'{num_val} {str_val}')
print(f'{num_val % 2 = }')
print(f'{str_val!r}')
price_val = 5.18362
print(f'{price_val:.2f}')
from datetime import datetime;
date_val = datetime.utcnow()
print(f'{date_val=:%Y-%m-%d}')

addresses = ["123 Elm Street", "531 Oak Street", "678 Maple Street"]
street = "Elm Street"
# 方法一
for address in addresses:
if address.find(street) >= 0:
print(address)
# 方法二
for address in addresses:
if street in address:
print(address)

str1 = "hello"
str2 = "😀"
def str_size(s):
return len(s.encode('utf-8'))
print(str_size(str1))
print(str_size(str2))

# 方法一
import os
exists = os.path.isfile('/path/to/file')
# 方法二
from pathlib import Path
config = Path('/path/to/file')
if config.is_file():
pass
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
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