统计字符串、九宫格、编码问题

统计字符串

while 1:
    number = char = space = other = 0
    strings = input("Please inpur a string(quit will be exit):")
    if strings == 'quit' or strings == 'exit':
        exit(1)
    for i in strings:
        if i.isdigit():
            number+=1
        elif i.isalpha():
            char+=1
        elif i.isspace():
            space+=1
        else:
            other+=1
    print('number={0}\nchar={1}\nspace={2}\nother={3}'.format(number,char,space,other))

练习2：
ABCD乘9=DCBA，A=? B=? C=? D=? 答案：a=1,b=0,c=8，d=9 1089*9=9801

for A in range(1,10):
for B in range(0,10):
for C in range(0,10):
for D in range(1,10):
if A1000+B100+C10+D==(D1000+C100+B10+A)9:
print('{0}{1}{2}{3}9={4}{5}{6}{7}'.format(A,B,C,D,D,C,B,A))

九宫格练习3：

number=list()
count=1
for i in range(10):
    number.append(i)
num=[(a,b,c) for a in number for b in number if a!=b for c in number if a!=c and b!=c and a+b+c==15]
for L1 in num:
    for L2 in num:
        if set(L1) & set(L2):
            continue
        for L3 in num:
            if set(L1) & set(L2) & set(L3):
                continue
            if L1[0]+L2[0]+L3[0]!=15:
                continue
            elif L1[1]+L2[1]+L3[1]!=15:
                continue
            elif L1[2]+L2[2]+L3[2]!=15:
                continue
            elif L1[0]+L2[1]+L3[2]!=15:
                continue
            elif L1[2]+L2[1]+L3[0]!=15:
                continue
            else:
                print('''
                                 第{9}种例子
                                 -------------
                                 | {0} | {1} | {2} |
                                 | {3} | {4} | {5} |
                                 | {6} | {7} | {8} |
                                 -------------'''.format(L1[0], L1[1], L1[2], L2[0], L2[1], L2[2], L3[0], L3[1], L3[2], count))
                count += 1

编码问题

只有在python2中才有编码问题
在python2默认编码是ASCII, python3里默认是unicode
unicode 分为 utf-32(占4个字节),utf-16(占两个字节)，utf-8(占1-4个字节)， so utf-16就是现在最常用的unicode版本， 不过在文件里存的还是utf-8，因为utf8省空间