我学习用python思考的复盘图

Aug 24, 2017

这里是我读Allen B. Downey的Think Python: How to Think Like a Computer Scientist作的复盘笔记，争取每天更新一章，估计四周读完第一遍。Mindnode的笔记还是原汁原味地贴图出来好看，最后我再导出Markdown格式，更新在这里。

整体图景

2017-07-26。

Wozniak建议没有整体图景之前不要拼图，这样效率相对低。我就把目录抄了一遍。果然感觉更舒服。

整体图景

0. 前言的话

2017-07-26.

前言的话

1. 程序语言

2017-07-26，这是对照原文写的，有过细之嫌。下次只写对自己有提示的地方就可以了。我不是做讲义，而是给自己做复盘。

程序语言

2. 变量陈述

2017-07-27，这次几乎是背着写的，有主动回忆，感觉挺好。

变量陈述

3. 函数

2017-07-28，今天学得不慢，做练习不太顺：E3我没有读清楚问题，自己给自己设定新任务：用一个参数控制打出多大的格子。但是这个功能如果不用循环语句，只用目前见过的命令，应该做不到。

4. 案例：交互界面设计

2017-07-31，读课文，做内部例题。我Debug的方式太傻了，其实一眼就该看出来sin函数给了负数值。
2017-08-01，做复盘，做课后练习。还有两道联系题，螺线和字母表，都应该很有意思，今天来不及做了。

5. 条件和反复

2017-08-02，末两个练习用到上一章turtle的练习，今天没做，也留着。

练习5.14-1：把以秒计的格林威治时间转为当前时间。

上午做了一小时，午饭后跑出来了。又写第二版，用recurse。

第一版：

import time
greenwich = time.time()

hash_one_year = 86400 * 365
hash_years = greenwich // hash_one_year

def that_year(n):
    """Calculate that nth year after 1970 has how many seconds

        n: the nth year after 1970
    """
    y = 1970 + n
    if (y % 4) == 0:
        that_year = 366 * 86400
    else:
        that_year = 365 * 86400
    return that_year


year_remain = greenwich
now_year = 1970
for i in range(int(hash_years)):
    if year_remain > 365*86400: # 365-There might be a little bug.
        year_remain = year_remain - that_year(i)
    now_year = 1970 + i + 1


def that_month(m, this_year):
    """Calculate that mth month has how many seconds

       m: the m th month
       this_year: this year
    """
    if (m==1) or (m==3) or (m==5) or(m==7) or(m==8) or(m==10) or(m==12):
        that_month = 31 * 86400
    elif m == 2:
        if this_year % 4 == 0:
            that_month = 29 * 86400
        else:
            that_month = 28 * 86400
    else:
        that_month = 30 * 86400
    return that_month


month_remain = year_remain
now_month = 1
for i in range(12):
    if month_remain > 28*86400:
        month_remain = month_remain - that_month(i+1, now_year)
        now_month = (i + 1) + 1


day_remain = month_remain
now_day = day_remain//86400

hour_remain = day_remain % 86400
now_hour = hour_remain // 3600

min_remain = hour_remain % 3600
now_min = min_remain // 60

sec_remain = min_remain % 60
now_sec = int(sec_remain)

print("year", now_year, "month:", now_month, "day:", now_day, "hour:", now_hour, "min:", now_min, "sec:", now_sec)

第一版结果。

year 2017 month: 8 day: 1.0 hour: 9.0 min: 42.0 sec: 19

我们比英国早八小时，在它的基础上+8，就是我的当下时间：2017-08-02 17:42:19

第二版代码：

import time
greenwichtime = time.time()

def that_year(n):
    """Calculate that nth year after 1970 has how many seconds

        n: the nth year after 1970
    """
    y = 1970 + n
    if (y % 4) == 0:
        that_year = 366 * 86400
    else:
        that_year = 365 * 86400
    return that_year


def that_month(m, this_year):
    """Calculate that mth month has how many seconds

       m: the m th month
       this_year: this year
    """
    if (m==1) or (m==3) or (m==5) or(m==7) or(m==8) or(m==10) or(m==12):
        that_month = 31 * 86400
    elif m == 2:
        if this_year % 4 == 0:
            that_month = 29 * 86400
        else:
            that_month = 28 * 86400
    else:
        that_month = 30 * 86400
    return that_month


def this_year(year_remain, k):
    now_year = 1970 + k
    year_remain = year_remain - that_year(k)
    print([year_remain, now_year])
    if year_remain < 0:
        year_remain = year_remain + that_year(k)
        return [year_remain, now_year]
    else:
        this_year(year_remain, k + 1)


now_year = this_year(greenwichtime, 0)[1]

print(now_year)

每年的数据都对，一直数到2017。但是函数返回值没有，只打印None。怎样用有结果的函数，怎样返回两个值，明天看下一章讲不讲。今天大致翻了一下，没有返回两个值的。在dash的python文档里搜function+return，没有谈这个问题。

...
[50059475.51102424, 2015]
[18437075.511024237, 2016]
[-13098924.488975763, 2017]
Traceback (most recent call last):
  File "/Users/johnqu/Projects/think_python3/code/greenwichtime_v2_john.py", line 46, in <module>
    now_year = this_year(greenwichtime, 0)[1]
TypeError: 'NoneType' object is not subscriptable

练习5.14-4，递归调用累计自然数相加

我写的docstring：

def recurse(n, s):
    """Calculate the integer n's addtion accumulation.

       eg. n = 5, recurse prints the result of 5 + 4 + 3 + 2 + 1
       thus:
       n: positive interger, if negative, it will exceed the maximum recursion depth around 992 extra.
       s: 0 is a prefered base. 
    """
    if n == 0:
        print(s)
    else:
        print("recurse", "n-->", n, "s-->", s)
        recurse(n-1, n+s)


recurse(5, 0)

如果用for语句写，可能是这个样子：

k = 0
for i in range(n)
    k = k + i + 1
print(k)

我是真的用3、用6演算过，从打印的状态表，才看出这函数在做加法。注意：print要在递归调用之前，打出来才是从定到底的状态表，否则是颠倒过来的。

recurse n--> 5 s--> 0
recurse n--> 4 s--> 5
recurse n--> 3 s--> 9
recurse n--> 2 s--> 12
recurse n--> 1 s--> 14
15

6. 有返回值的函数

2017-08-03，读完做完。今天就不贴代码，对比经验写在下图中了。（貌似越写越多）

7. 反复

2017-08-04。算法的书，资源，打住。

8. 字符串

2017-08-07，明天要做文字游戏了，大战一场。

Keep that in mind in case you have to defend yourself against a man armed with a Pineapple.啥意思？

9. 案例：文字游戏

2017-08-08做题，没来得及复盘。

10. 清单

2017-08-09。习题做了8个，还有四个来不及做。

11. 字典

2017-08-10，只做了2/6道题。

2017-08-10。

The in operator works on dictionaries, too; it tells you whether something appears as a key in the dictionary (appearing as a value is not good enough).

为什么不够好？

For dictionaries, Python uses an algorithm called a hashtable that has a remarkable property: the in operator takes about the same amount of time no matter how many items are in the dictionary. I explain how that’s possible in Section B.4, but the explanation might not make sense until you’ve read a few more chapters.

将来要读读。

"""E11-2 2017-08-10
"""

def histogram(s):
    """E11-2: rewrite the histogram with .get()method

    s: string
    return: dist of charactor and counter
    """
    d = dict()
    for c in s:
        d[c] = d.get(c, 0) + 1
    return d

h = histogram('brontosaurus')
print(h)

E11-10-2

inverse.setdefault(val, [key]).append(key产生的结果是：