List和Dictionary 先举两个Dictionary的例子

prices={"banana":4, "apple":2, "orange":1.5, "pear":3}

lloyd={"name":"Lloyd", "homework":[], "quizzes":[], "tests":[]}

引用”banana”价格时为prices[“banana”]

再举一个List的例子

animals=["monkey", "dog", "cat"]

List和Dictionary都可用for遍历

Removing elements from lists

n.pop(index) will remove the item at index from the list and return it to you:

n = [1, 3, 5]
n.pop(1)
# Returns 3 (the item at index 1)
print n
# prints [1, 5]

n.remove(item) will remove the actual item if it finds it:

n.remove(1)
# Removes 1 from the list,
# NOT the item at index 1
print n
# prints [3, 5]

del(n[1] is like .pop in that it will remove the item at the given index, but it won’t return it:

del(n[1])
# Doesn't return anything
print n
# prints [1, 5]

Arbitrary number of arguments

def add_function(*args)

is the convention for allowing an arbitrary number of arguments. *args is an argument tuple that can be called the same way you would a list! For example:

def myFun(*args):
    print args[0]

The above example would print out the first argument that was given.

To sum those args, you can just use Python’s built-in sum function: sum(args) (no * needed when calling sum!)

Using an arbitrary number of lists in a function

m = [1, 2, 3]
n = [4, 5, 6]
# Add your code here!
def join_lists(*args):
f=""
for s in args:
f+=s
print f

print join_lists(m, n)

Traceback (most recent call last): File “1.py”, line 15, in print join_lists(m, n) File “1.py”, line 8, in join_lists f+=s TypeError: cannot concatenate ‘str’ and ‘list’ objects

原因是list只能和list连接 将f=""改成f=[]就可以了! 疑惑:为什么list+list不可以用sum()

函数中使用for循环遍历数组时函数最后使用return返回并用print显示可能只显示第一个元素

board = []

for x in range(0, 5):
    board.append(["O"] * 5)

def print_board(b):
    for row in b:
        return " ".join(row)
print print_board(board)

结果为0 0 0 0 0

此时将函数中 return ” “.join(row) 改为 print ” “.join(row)

即可正常显示所需结果: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

for语句中else的意义

In this case, the else statement is executed after the for, but only if the for ends normally—that is, not with a break.This code will break when it hits ‘tomato’, so the **else **block won’t be executed.

fruits = ['banana', 'apple', 'orange', 'tomato', 'pear', 'grape']

print 'You have...'
for f in fruits:
    if f == 'tomato':
        print 'A tomato is not a fruit!' # (It actually is.)
        break
    print 'A', f
else:
    print 'A fine selection of fruits!'

此处else的意义在于在所有遍历结束后可以只打印一句话，且这句话可以在for结构内

factorial

我想到的方法

    def factorial(x):
        if x==0:
            f=0
        else:
            f=1
        while x>1:
            f=x*f
            x-=1
        else:
            print f
    factorial(3)

网上找到的好方法

    def factorial(x):
        if x == 1:
            return x
        else:
            return x * factorial(x-1)

for遍历输出到一行

def reverse(text):
    rev=""
    for i in text:
        rev=i+rev
    return rev