问题描述

如 PythonCookbook 中所述，可以在元组之前添加 *.* 在这里是什么意思?

<块引用>

第 1.18 章.将名称映射到序列元素:

从集合导入命名元组Stock = namedtuple('Stock', ['name', 'shares', 'price'])s = 库存(*rec)# 这里rec是一个普通的元组，例如:rec = ('ACME', 100, 123.45)

在同一部分，**dict 呈现:

<块引用>

从集合导入命名元组Stock = namedtuple('Stock', ['name', 'shares', 'price', 'date', 'time'])# 创建原型实例stock_prototype = Stock('', 0, 0.0, None, None)# 将字典转换为股票的函数def dict_to_stock(s):return stock_prototype._replace(**s)

**在这里的作用是什么?

解决方案

在函数调用中

*t 表示将此可迭代对象的元素视为此函数调用的位置参数."

def foo(x, y):打印(x，y)>>>t = (1, 2)>>>脚)1 2

从 v3.5 开始，您还可以在列表/元组/集合文字中执行此操作:

>>>[1, *(2, 3), 4][1、2、3、4]

**d 表示将字典中的键值对视为此函数调用的附加命名参数."

def foo(x, y):打印(x，y)>>>d = {'x':1, 'y':2}>>>食物)1 2

从 v3.5 开始，您还可以在字典文字中执行此操作:

>>>d = {'a': 1}>>>{'b': 2, **d}{'b':2，'a':1}

在函数签名中

*t 表示将所有附加的位置参数作为一个元组打包到这个参数中."

def foo(*t):打印(t)>>>富(1, 2)(1, 2)

**d 表示将所有附加的命名参数带到此函数，并将它们作为字典条目插入到此参数中."

def foo(**d):打印(d)>>>富(x=1，y=2){'y':2，'x':1}

在赋值和 for 循环中

*x 表示消耗右侧的附加元素"，但它不必是最后一项.请注意，x 将始终是一个列表:

>>>x, *xs = (1, 2, 3, 4)>>>X1>>>xs[2, 3, 4]>>>*xs, x = (1, 2, 3, 4)>>>xs[1、2、3]>>>X4>>>x, *xs, y = (1, 2, 3, 4)>>>X1>>>xs[2, 3]>>>是的4>>>对于 [ (1, 2, 3, 4) ] 中的 (x, *y, z): print(x, y, z)...1 [2, 3] 4

---

请注意，出现在 * 之后的参数仅限关键字:

def f(a, *, b): ...f(1, b=2) # 很好f(1, 2) # 错误:b 仅是关键字

Python3.8 添加了仅位置参数, 表示不能用作关键字参数的参数.它们出现在 / 之前(* 之前的关键字参数的双关语).

def f(a,/, p, *, k): ...f( 1, 2, k=3) # 很好f( 1, p=2, k=3) # 很好f(a=1, p=2, k=3) # error: a is positional-only

As mentioned in PythonCookbook, * can be added before a tuple. What does * mean here?

Chapter 1.18. Mapping Names to Sequence Elements:

from collections import namedtuple
Stock = namedtuple('Stock', ['name', 'shares', 'price'])
s = Stock(*rec) 
# here rec is an ordinary tuple, for example: rec = ('ACME', 100, 123.45)

In the same section, **dict presents:

from collections import namedtuple
Stock = namedtuple('Stock', ['name', 'shares', 'price', 'date', 'time'])
# Create a prototype instance
stock_prototype = Stock('', 0, 0.0, None, None)
# Function to convert a dictionary to a Stock
def dict_to_stock(s):
    return stock_prototype._replace(**s)

What is **'s function here?

解决方案

In a function call

*t means "treat the elements of this iterable as positional arguments to this function call."

def foo(x, y):
    print(x, y)

>>> t = (1, 2)
>>> foo(*t)
1 2

Since v3.5, you can also do this in a list/tuple/set literals:

>>> [1, *(2, 3), 4]
[1, 2, 3, 4]

**d means "treat the key-value pairs in the dictionary as additional named arguments to this function call."

def foo(x, y):
    print(x, y)

>>> d = {'x':1, 'y':2}
>>> foo(**d)
1 2

Since v3.5, you can also do this in a dictionary literals:

>>> d = {'a': 1}
>>> {'b': 2, **d}
{'b': 2, 'a': 1}

In a function signature

*t means "take all additional positional arguments to this function and pack them into this parameter as a tuple."

def foo(*t):
    print(t)

>>> foo(1, 2)
(1, 2)

**d means "take all additional named arguments to this function and insert them into this parameter as dictionary entries."

def foo(**d):
    print(d)

>>> foo(x=1, y=2)
{'y': 2, 'x': 1}

In assignments and for loops

*x means "consume additional elements in the right hand side", but it doesn't have to be the last item. Note that x will always be a list:

>>> x, *xs = (1, 2, 3, 4)
>>> x
1
>>> xs
[2, 3, 4]

>>> *xs, x = (1, 2, 3, 4)
>>> xs
[1, 2, 3]
>>> x
4

>>> x, *xs, y = (1, 2, 3, 4)
>>> x
1
>>> xs
[2, 3]
>>> y
4

>>> for (x, *y, z) in [ (1, 2, 3, 4) ]: print(x, y, z)
...
1 [2, 3] 4

---

Note that parameters that appear after a * are keyword-only:

def f(a, *, b): ...

f(1, b=2)  # fine
f(1, 2)    # error: b is keyword-only

Python3.8 added positional-only parameters, meaning parameters that cannot be used as keyword arguments. They appear before a / (a pun on * preceding keyword-only args).

def f(a, /, p, *, k): ...

f(  1,   2, k=3)  # fine
f(  1, p=2, k=3)  # fine
f(a=1, p=2, k=3)  # error: a is positional-only