在这种情况下，您能帮我吗？ 我有3个清单。
list1 = [ "a", "b", "c", "d"]
list2 = ["x", "y", "z"]
list3 = [1, 2, 3, 4, 5, 6, 7]
首先，我需要获取list1的3个元素，list2的2个元素和list3的4个元素，然后将它们组合起来：
[a, b, c, x, y, 1, 2, 3, 4]
[a, b, c, x, z, 1, 2, 3, 5]
[a, b, c, x, z, 1, 2, 3, 6]
[a, b, c, x, z, 1, 2, 3, 7]
....
然后我需要在末尾打印出list3其余部分的3个元素：
[a, b, c, x, y, 1, 2, 3, 4, 5, 6, 7]
[a, b, c, x, z, 1, 2, 3, 5, 4, 6, 7]
[a, b, c, x, z, 1, 2, 3, 6, 4, 5, 7]
[a, b, c, x, z, 1, 2, 3, 7, 4, 5, 6]
....
我使用了2个列表的减法，但显示错误
TypeError: unsupported operand type(s) for -: 'list' and 'itertools.combinations'
...
combi1 = combinations(list1, 3)
combi2 = combinations(list2, 2)
combi3 = combinations(list3, 4)
rest3 = list3 - combi3
com = product(combi1, combi2, combi3)
请帮助我解决这个问题！ 非常感谢！
python
答案 0 :(得分：0)
您可以使用import ast, inspect
import numpy as np
import sympy as sp
def f(a):
return np.array([np.sin(a), np.cos(a)])
def f2(a):
return np.array([1, np.sin(a)])
def f3(a):
return f(a) + f2(a)
translate = {'array': 'Array'}
class np_to_sp(ast.NodeTransformer):
def visit_Name(self, node):
if node.id=='np':
node = ast.copy_location(ast.Name(id='sp', ctx=node.ctx), node)
return node
def visit_Attribute(self, node):
self.generic_visit(node)
if node.value.id=='sp' and node.attr in translate:
fields = {k: getattr(node, k) for k in node._fields}
fields['attr'] = translate[node.attr]
node = ast.copy_location(ast.Attribute(**fields), node)
return node
from types import FunctionType
for fn in f3.__code__.co_names:
fo = globals()[fn]
if not isinstance(fo, FunctionType):
continue
z = ast.parse(inspect.getsource(fo))
np_to_sp().visit(z)
exec(compile(z, '', 'exec'))
x = sp.Symbol('x')
print(f3(x))
来对[sin(x) + 1, sin(x) + cos(x)]
之后的元组列表进行展平，要获取其余列表，可以使用chain-product（注意，这需要您的list1到list3没有重叠）
set