数据类型-习题解答

1.简述变量名规范:

  字母数字下划线,不能以数字开头

2.字节占位的关系

  一字节等于八位

3.utf-8占3个字节,24位。gbk占2个字节,16位

4.字符串的12个功能:

  1.len() 长度

  2.“+” 拼接

  3.upper()大写

  4.count() 指定字符出现次数

  5.split() 分割

  6.index()返回索引值

  7.find()返回索引值

  8.join() 拼接

5.数字,字符串,列表,元组,字典对应的布尔值False分别是:0,空值

6.py2与py3区别

  print输出

  字符编码格式

7.li = [1,3,2,“a”,"4","b",5,"c"]

  1).形成列表[1,2,4,5]-------li[ : : 2]

  2.

8:

lis = [["k",["qwe",20,{"k1":["tt",3,"1"]},89],"ab"]]
# 将“tt”大写
#第一种方法
# lis[0][1][2]["k1"] = ["TT",3,"1"]
# print(lis)
#第二种方法
# list(lis[0][1][2].values())[0][0]=list(lis[0][1][2].values())[0][0].upper()
# print(lis)
#将3变成字符串“100”
#第一种方法
# lis[0][1][2]["k1"] = ["tt","100","1"]
# print(lis)
#第二种方法
# lis[0][1][2]["k1"][1]="100"
# print(lis)

#将"1"变成101
lis[0][1][2]["k1"] = ["tt",3,101]
print(lis)

 9:

dic = {"k1":"v1","k2":["alxe","sb"],(1,2,3,4,5):{"k3":["2",100,"wer"]}}
#将k2对应的值后面添加元素“23”
# dic["k2"].append("23")
# print(dic)

#将k2的第一个值添加元素“a”
# dic["k2"].insert(0,"a")
# print(dic)

#将(1,2,3,4,5)对应的值添加一个键值对"k4":"v4"
# dic[(1,2,3,4,5)]["k4"]="v4"
# print(dic)

#将k3对应值“wer”改为qq
#dic[(1,2,3,4,5)["k3"]].replace("wer","qq")
# dic[(1,2,3,4,5)]["k3"][2]= 'qq'
# print(dic)

 10:

#将列表构造成字符串
# li = ["ss","d","ddsw"]
#s = "".join(li)

#构造成字符串s = "ss*d*ddsw"
# s = "*".join(li)
# print(s)

#有字符串s = "ssdadsw",构造成列表li = ["sd","dsw"]
# s = "ssdadsw"
# li = s.split("a")
# print(li)

#有字符串s = "ssda dsw",构造成列表li = ["ssda","dsw"]
# s = "ssda dsw"
# li = s.split(" ")
# print(li)

#s= "alse" 构造字符串s1 = "a_l_s_e"
# s= "alse"
# s1 = "_".join(s)
# print(s1)

10:

#分别使用for ,while打印1-2+3-4...+99
sum = 0
# for i in range (1,100):
#     if i % 2 == 0:
#         sum -= i
#     else:
#         sum += i
# print(sum)

i = 1
while i <100:
    if i % 2 == 0:
        sum -= i
    else:
        sum += i
    i += 1
    if i == 100:
        break
print(sum)

 11.

# 使用range()打印100,99,...0
li = []
# for i in range (0,101):
#     li.append(100-i)
# print(li)

# for i in range (0,101):
#      li.append(i)
#      li.reverse()
# print(li)

12.

#计算用户输入内容中索引为奇数并且对应元素为数字个数(没有则数字为0)
li = input("请输入任意内容:")
j = 0
for i in range(len(li)):
    if i % 2 != 0:
       # print(li[i])
        if li[i].isdigit():
            print(li[i])
            j += 1
print("数字个数为%d"%j)

 13.

#将li中大于55的为值,小于55的为键
li = [11,22,33,44,66,77,88]
dic = {}
key = []
val = []
for row in li:
    if row < 55:
        key.append(row)
    else:
        val.append(row)
key = tuple(key)      #字典键为不能修改,所以设置为元组
print(key)
print(val)
dic[key] = val
print(dic)

 14:

#。查找列表li中的元素,移除每个元素的空格,并找出以"A""a"开头,并以"c"结尾的元素并添加到列表中循环打印

li = ["taibao","aldfC","Ak;dkc","aldc","AkkdfC","SDL","  ADDc"]
resu = []
for i in li:
    row = i.strip()  #移除空格
    if row[0].upper() == "A" and row[-1] == "c":
        print(row)
        resu.append(row)
print(resu)

15;

#计算器
content = input("请输入内容:")
li = content.split("+")
sum = 0
for i in li:
    sum += int(i)
print("%s=%d"%(content,sum))
原文地址:https://www.cnblogs.com/wzc27229/p/10970725.html