<a href="http://www.jianshu.com/p/54870e9541fc">总目录</a>
课程页面：https://www.codecademy.com/
内容包含课程笔记和自己的扩展折腾
is_even
def is_even(n):
if n % 2 == 0:
return True
else:
return False
print is_even(16)
Output:
True
is_int
def is_int(n):
if n == int(n):
return True
else:
return False
"""
这里之所以不用type(n),
是因为python2中，你问他7.0的type是啥
python会说是float
但是一般人类都默认7.0是integer
"""
print is_int(7.0)
Output:
True
digit_sum
给一个多位的数字，return各个位数之和，eg. 给1234 return 10
def digit_sum(n):
n = str(n)
result = 0
for i in range(len(n)):
result += int(n[i])
return result
print digit_sum(666)
Output:
18
factorial
ps 这里的阶乘只针对非负数的整数哦，且0！= 1
def factorial(n):
result = 1
if n == 0:
return 1
elif n > 0 and int(n) == n:
for i in range(1, n+1):
result *= i
return result
else:
return 'null'
print factorial(0)
print factorial(3)
print factorial(3.5)
print factorial(-3.5)
Console:
1
6
null
null
is_prime
A prime number is a positive integer greater than 1 that has no positive divisors other than 1 and itself. 中文：素数/质数，例如：2，3，5，7...
【方法一：用for loop，比较麻烦】
def is_prime(n):
if n < 2 or int(n) != n:
return False
elif n == 2:
return True
else:
for i in range(2, n):
if n % i == 0:
return False
break
else:
return True
print is_prime(1)
print is_prime(2)
print is_prime(3)
print is_prime(5)
print is_prime(222)
print is_prime(-9)
print is_prime(0)
print is_prime(2.5)
Console:
False
True
True
True
False
False
False
False
Process finished with exit code 0
【方法二：用while loop，代码会少一些】
def is_prime(n):
a = 2
if n < 2 or int(n) != n:
return False
else:
while a < n-1:
if n % a == 0:
return False
break
a += 1
else:
return True
print is_prime(1)
print is_prime(2)
print is_prime(3)
print is_prime(5)
print is_prime(222)
print is_prime(-9)
print is_prime(0)
print is_prime(2.5)
Console:
False
True
True
True
False
False
False
False
Process finished with exit code 0
Modifying strings
（从我自己的2. Strings & Console Output复制过来的）
【自创练习1：把string中的a/A全部换成*】
def no_a(my_string):
for i in range(len(my_string)):
if my_string[i] == "a" or my_string[i] == "A":
my_string[i] = "*" # error!
return my_string
print no_a("Happy birthday!")
Output:
Traceback (most recent call last):
File "/Users/y(***)zhang/PycharmProjects/untitled1/test.py", line 11, in <module>
print no_a("Happy birthday!")
File "/Users/y(***)zhang/PycharmProjects/untitled1/test.py", line 8, in no_a
my_string[i] = "*"
TypeError: 'str' object does not support item assignment
Process finished with exit code 1
搜了一下发现：strings are immutable... 心碎
看来只能曲线救国了：
【方法一：】strings变成list，至少strings之间能concatenate，不算太差
def no_a(my_string):
result = []
for i in range(len(my_string)):
if my_string[i] != "a" and my_string[i] != "A":
result.append(my_string[i])
else:
result.append("*")
for i in range(1, len(result)):
result[0] = result[0] + result[i]
return result[0]
"""
上面可以多放个变量，
比如new_string = "",
然后for loop里面不从1开始从0开始，
最后return new_string.
不过开始写程序的时候我懒了。（结果现在还要多打说明comment...)
"""
print no_a("Happy birthday!")
Console:
H*ppy birthd*y!
Process finished with exit code 0
【方法二：】用自带的.join(list)的功能，更加简单
def no_a(my_string):
result = []
for i in range(len(my_string)):
if my_string[i] != "a" and my_string[i] != "A":
result.append(my_string[i])
else:
result.append("*")
"""
a = "".join(result)
print type(a)
结果是，join之后的格式是string: <type 'str'>
"""
return "".join(result)
print no_a("Happy birthday!")
Console:
H*ppy birthd*y!
Process finished with exit code 0
reverse
给一个string，譬如"zy"，把反过来变成"yz"
【方法一：简单方法】
def reverse(ms):
result = ""
for char in ms:
result = char + result
return result
print reverse("Happy")
【方法二：用range，复习一下range的step是负数的功能】
def reverse(ms):
n = len(ms)
result = ""
for i in range(n-1, -1, -1):
result = result + ms[i]
return result
print reverse("Happy")
Console:
yppaH
Process finished with exit code 0
anti_vowel
anti_vowel("Hey You!") should return "Hy Y!"
【方法一：我的偷懒做法】
def anti_vowel(ms):
result = ""
for char in ms:
if char not in "aeiouAEIOU":
result += char
return result
print anti_vowel("Happy birthday")
Output:
Hppy brthdy
【方法二：codecademy的答案】
def anti_vowel(text):
vowels = ["a", "e", "i", "o", "u"]
result = ""
for char in text:
test = char.lower()
if test not in vowels:
result = result + char
return result
scrabble_score
给一个dictionary, 里面的tuple是：key-字母，value-数值。然后输入字母，输出这些字母对应的数值。
score = {
"a": 1, "c": 3, "b": 3, "e": 1, "d": 2, "g": 2,
"f": 4, "i": 1, "h": 4, "k": 5, "j": 8, "m": 3,
"l": 1, "o": 1, "n": 1, "q": 10, "p": 3, "s": 1,
"r": 1, "u": 1, "t": 1, "w": 4, "v": 4, "y": 4,
"x": 8, "z": 10
}
def scrabble_score(word):
result = 0
word = word.lower()
for char in word:
result += score[char]
return result
print scrabble_score("Abc")
Output:
7
censor
要达到如下效果：
censor("this hack is wack hack", "hack")
输出："this **** is wack ****"
def censor(text, word):
text = text.split()
for i in range(len(text)):
if text[i] == word:
text[i] == "*"*len(text[i])
text = " ".join(text)
return text
print censor("this hack is wack hack", "hack")
Output:
this hack is wack hack
remove_duplicates
输入[4, 1, 2, 2, 2, 1, 3], 输出[1, 2, 3, 4]
def remove_duplicates(ml):
result = []
for i in range(len(ml)):
if ml[i] not in result:
result.append(ml[i])
return sorted(result)
print remove_duplicates([4, 1, 2, 2, 2, 1, 3])
Output:
[1, 2, 3, 4]
median
输入[4, 1, 2, 2, 2, 1, 3] 输出 2
输入[1, 1, 2, 2] 输出 1.5
def median(ml):
ml = sorted(ml)
n = len(ml)
if n % 2 == 0:
return 0.5*(ml[n/2-1] + ml[n/2+1-1])
else:
return ml[(n+1)/2-1]
print median([4, 1, 2, 2, 2, 1, 3])
print median([1, 1, 2, 2])
# 注意，计算机从0开始，统计学从1开始
Output:
2
1.5
