用python解决这些问题，每道题5元，见代码给

Python (BSU FAMCS Fall’19)
Seminar 3
Advisor: Dzmitryi Kasitsyn
Task 1. (0.5 points). Implement a function transpose to transpose an iterable of nested iterables.
All nested iterables are supposed to have the same length. In other words you’re asked to transpose a
two-dimensional rectangular array.
Consider to use built-in functions and functions from itertools module. Don’t use any loops and
do not explicitly convert the given iterable to a list or to a tuple implementing your solution.
Save your function into a iter_helpers.py file.
Example

expected = [[1 , 2], [-1 , 3]]
actual = transpose ([[1 , -1], [2 , 3]])
assert expected == list (map(list , actual ) )

Task 2. (1 point). Implement a function scalar_product to calculate a scalar product of two
iterables that can contain items of type int, float, or str.
String items can contain:
∙ an integer representation (binary, octal, decimal, hexadecimal with corresponding prefixes) – use
built-in int function to convert them to a number,
∙ some letters that cannot be interpreted as numbers – assume the result of the whole expression
is None in that case.
Consider to use built-in functions and functions from itertools module. Don’t use any loops
implementing your solution.
Save your function into a iter_helpers.py file.
Example

expected = 1
actual = scalar_product ([1 , ’2’], [-1 , 1])
assert expected == actual
actual = scalar_product ([1 , ’xyz ’], [-1 , 1])
assert actual is None

Task 3. (0.5 points). Implement a decorator profile that calculates how long a given function is
being executed, prints the execution time and returns the function result.
Consider to use timeit module and default_timer function to calculate function execution time.
Save your decorator into a utils.py file.
Example

@profile
def some_function ():
return sum ( range ( 1000 ) )
result = some_function () # return a value and print execution time

Task 4. (0.5 points). Implement a context manager timer to calculate code block execution time
and print it onto a screen.
Save the context manager into a utils.py file.
Example

with timer ():
print ( sum ( range ( 1000 ) ) )
# print execution time when calculation is over
1

Task 5. (1.5 points). Suppose you are given a singly linked list node class Node (see below) so that
a signly linked list is a sequence of Nodes that have either other Node or None as node’s next_ value.
The value None determines the end of a list.
Nested lists have other Nodes stored in value field.
You are asked to do the next:
∙ Add check of argument types in the constructor (init method; use assert);
∙ Implement properties to get and set _value and _next values;
∙ Implement a «magic» (dunder ) method iter to iterate over a list;
∙ Implement a flatten_linked_list function to flatten your list inplace. First, it has to modify
a given list in a way that all _values don’t have Node type however the sequence of values is
the same as in the original list iterating over it. Second, your function have to return the object
passed into it (refer to r3 example below).
Save your solution in a linked_list.py file.
Example

[](class Node ( object ):
def __init__ ( self , value , next_ = None ):
self . _value = value
self . _next = next_
r1 = Node ( 1 ) # 1 -> None - just one node
r2 = Node (7 , Node (2 , Node ( 9 ) ) ) # 7 -> 2 -> 9 -> None
# 3 -> (19 -> 25 -> None ) -> 12 -> None
r3 = Node (3 , Node ( Node ( 19 , Node ( 25 ) ) , Node ( 12 ) ) )
r3_flattenned = flatten_linked_list ( r3 ) # 3 -> 19 -> 25 -> 12 -> None
r3_expected_flattenned_collection = [3 , 19 , 25 , 12]
assert r3_expected_flattenned_collection == list ( r3_flattenned )

第二题

import numpy as np
def scalar_product(a_array,b_array):
    result=None
    try:
        a_array=np.array(a_array,dtype=int)
        #转换为整型
        b_array=np.array(b_array,dtype=int)
        result=sum(a_array*b_array)
        #数组对应相乘再求和，其实也可以用我之前说的切片的方法
        #result=np.dot(a,b)#这是numpy自带的点积函数
    except:
        print('不满足转换条件')
    print(result if result!=None else '内积失败')
scalar_product([1,'1'],[2, 2])
scalar_product([1,'az'],[2,2])

两种运行结果

img

Task 1:

img

def transpose(data):
    return zip(*data)


if __name__ == '__main__':
    data = [[1, 2, 4], [-1, 3, 5], [6, 7, 8]]
    result = transpose(data)
    print(list(result))  # 验证


import numpy as np
def transpose(a_array):
    a_array=np.array(a_array)
    #不用numpy将其转换为数组的话，使用切片时，系统会把它当作元组而报错
    des_array=np.zeros_like(a_array)#创建一个和a_array一样的数组
    des_array[0,:]=a_array[:,0]
    #其实就是每个嵌套列表的对应位置提取出来形成一个新列表
    #若果说不止二维的话，那就在切片里做修改，多加几个，并调整就行
    des_array[1,:]=a_array[:,1]
    return des_array
actual=transpose([[1,-1],[2,3]])
expected=[[1,2],[-1,3]]
print(actual)
print((actual==expected).all())#判断书否相等并返回一个逻辑值True/False

img

img