第一部分：基础练习
1.1 请完成代码，实现输入任意多个空格分隔的浮点数，并打印输出这些浮点数的和
参考：
a, b = map(int, input('输入空格分隔的两个数：').split())
print(a+b)
提示：同时赋值两个变量的这个语法叫 Tuple Unpacking
#map和sum的用法: sum(list(map(float, inp.split())))
def do_sum(inp):
# 补充完成函数 Write here
return sum(list(map(float, inp.split())))
# 下面的语句用来检验 do_sum 的正确性
assert do_sum('1 2') == 3
assert do_sum('1 2 3 4.6 5 -1.1') == 14.5
assert do_sum('-123') == -123
print('Test passed!')
Test passed!
1.2 定义一个函数func(listinfo) listinfo:为列表，listinfo = [133, 88, 24, 33, 232, 44, 11, 44]，剔除列表中大于100的奇数
#filter和lambda的用法: list(filter(lambda x: x <= 100 or x % 2 == 0, listinfo))
def func(listinfo):
#TODO
return list(filter(lambda x: x <= 100 or x % 2 == 0, listinfo))
print(func([133, 88, 24, 33, 232, 44, 11, 44]))
[88, 24, 33, 232, 44, 11, 44]
1.3 牛顿迭代法
以方程：x ^3-0.2x ^2-0.2x-1.2=0为例，编写程序求该方程的根,精度为0.00000001
#牛顿迭代法: temp = temp - tempval / getdao(a, temp)
#浮点数比较大小: math.isclose(tempval, 0.0)
#保留位数: print("%.8f" % getres([-1.2, -0.2, -0.2, 1]))
import math
#求函数值
def getval(a, x):
r = 0.0
b = reversed(a)
for element in b:
r = r * x + element
return r
#求导
def getdao(a, x):
b = []
for i in range(1, len(a)):
b.append(a[i] * i)
return getval(b, x)
#牛顿迭代
def getres(a):
temp = 0.0;
tempval = getval(a, temp)
while not math.isclose(tempval, 0.0):
temp = temp - tempval / getdao(a, temp)
tempval = getval(a, temp)
return temp
print("%.8f" % getres([-1.2, -0.2, -0.2, 1]))
1.20000000
1.4 设计一个decorator，它可作用于任何函数上，并打印该函数的执行时间
#时间长: time.process_time()
#函数装饰器
import time
def biubiu(func):
def inner():
print("------decorate-------")
start = time.process_time()
func()
end = time.process_time()
print(str(end - start) + " seconds are used.")
return
return inner
@biubiu
def a():
print("i am a")
return
@biubiu
def b():
print("i am b")
return
a()
b()
------decorate-------
i am a
4.7999999999603915e-05 seconds are used.
------decorate-------
i am b
4.6000000001100716e-05 seconds are used.
1.5 用匿名函数改造下面的代码
def is_odd(n):
return n % 2 == 1
L = list(filter(is_odd, range(1, 20)))
print(L)
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]
#TODO
L = list(filter(lambda x: x % 2 == 1, range(1, 20)))
print(L)
[1, 3, 5, 7, 9, 11, 13, 15, 17, 19]
第二部分：进阶练习
2.1 给定一个数组nums找最长递增子序列长度（提示：用装饰器实现动态规划算法的记忆存储）
#动态规划和函数装饰器
def memorize(func):
memory = {}
def inner(a, i):
if i not in memory:
memory[i] = func(a, i)
return memory[i]
return inner
@memorize
def findres(a, i):
if i == 0:
return 1
else:
temp = 0
for j in range(0, i):
ttemp = findres(a, j)
if a[i] > a[j] and ttemp > temp:
temp = ttemp
return temp + 1
def lis(nums):
#TODO
temp = 0
for i in range(0, len(nums)):
ttemp = findres(nums, i)
if ttemp > temp:
temp = ttemp
return temp
print(lis([7,6,1,5,4,3,2,3,4,2,5]))
5
2.2 从开发的代码库中得到一组数据，表示每个文件的代码变更情况
{'login.py': 'a 8 d 2 u 3', 'order.py': 'a 15 d 0 u 34', 'info.py': 'a 1 d 20 u 5'}
其中 a表示新增行数，d表示删除行数，u表示修改行数。login.py的变更行数为13
要求：统计出每个文件的变更行数(提示：通过try方法)
s = {'login.py': 'a 8 d 2 u 3', 'order.py': 'a 15 d 0 u 34', 'info.py': 'a 1 d 20 u 5'}
res = {}
for k, v in s.items():
try:
v = v.split(" ")
n = int(v[1]) + int(v[3]) + int(v[5])
res[k] = n
except:
print("exception!")
print(res)
{'login.py': 13, 'order.py': 49, 'info.py': 26}
2.3 运行下面的代码，根据异常信息进行分析，定位出错误源头，并修复
#请看下一个代码块
#小数字符串转换成整数
#把int(s)改成了float(s)
from functools import reduce
def str2num(s):
return int(s)
def calc(exp):
ss = exp.split('+')
ns = map(str2num, ss)
return reduce(lambda acc, x: acc + x, ns)
def main():
r = calc('100 + 200 + 345')
print('100 + 200 + 345 =', r)
r = calc('99 + 88 + 7.6')
print('99 + 88 + 7.6 =', r)
main()
100 + 200 + 345 = 645
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-41-ad8040b068b8> in <module>
15 print('99 + 88 + 7.6 =', r)
16
---> 17 main()
<ipython-input-41-ad8040b068b8> in main()
12 r = calc('100 + 200 + 345')
13 print('100 + 200 + 345 =', r)
---> 14 r = calc('99 + 88 + 7.6')
15 print('99 + 88 + 7.6 =', r)
16
<ipython-input-41-ad8040b068b8> in calc(exp)
7 ss = exp.split('+')
8 ns = map(str2num, ss)
----> 9 return reduce(lambda acc, x: acc + x, ns)
10
11 def main():
<ipython-input-41-ad8040b068b8> in str2num(s)
2
3 def str2num(s):
----> 4 return int(s)
5
6 def calc(exp):
ValueError: invalid literal for int() with base 10: ' 7.6'
from functools import reduce
def str2num(s):
return float(s)
def calc(exp):
ss = exp.split('+')
ns = map(str2num, ss)
return reduce(lambda acc, x: acc + x, ns)
def main():
r = calc('100 + 200 + 345')
print('100 + 200 + 345 =', r)
r = calc('99 + 88 + 7.6')
print('99 + 88 + 7.6 =', r)
main()
100 + 200 + 345 = 645.0
99 + 88 + 7.6 = 194.6
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