第1关 列表的属性与方法
def main(n):
ls = []
for i in range(n):
lsinput = input().split()
operate = lsinput[0]
if operate == 'insert':
ls.insert(int(lsinput[1]),int(lsinput[2]))
elif operate == 'remove':
ls.remove(int(lsinput[1]))
elif operate == 'append':
ls.append(int(lsinput[1]))
elif operate == 'sort':
ls.sort()
elif operate == 'pop':
ls.pop()
elif operate == 'reverse':
ls.reverse()
elif operate == 'print':
print(ls)
if __name__ == '__main__':
N = int(input())
main(N)第2关 推导式与生成器
ls = ['the lord of the rings','anaconda','legally blonde','gone with the wind']
s = input() # 输入一个字符
# 当输入为"1"时，输出元素为0-9的3次方的列表 [0, 1, 8, 27, 64, 125, 216, 343, 512, 729]
if s == '1':
print([x ** 3 for x in range(10)])
# 当输入为"2"时，输出元素为0-9中偶数的3次方的列表 [0, 8, 64, 216, 512]
elif s == '2':
print([x ** 3 for x in range(10) if x % 2 == 0])
# 当输入为"3"时，输出元素为元组的列表，元组中元素依次是0-9中的奇数和该数的3次方[(1, 1), (3, 27), (5, 125), (7, 343), (9, 729)]
elif s == '3':
print([(x,x ** 3) for x in range(10) if x % 2 == 1])
# 当输入为"4"时，将ls中每个元素单词首字母大写输出['The lord of the rings', 'Anaconda', 'Legally blonde', 'Gone with the wind']
elif s == '4':
print([s.capitalize() for s in ls])
# 当输入为其他字符时，执行以下语句
else:
print("结束程序")
第3关 列表的合并与排序
lst_a = list(map(int,input().split()))
lst_b = list(map(int,input().split()))
lst_ab = lst_a + lst_b
lst_ab.sort(reverse=True)
print(lst_ab)第4关 二维列表排序
m = int(input())
n = int(input())
myList = [('dungeon',7),('winterfell',4),('bran',9),('meelo',6)]
if m > len(myList):
m = len(myList)
print(sorted(myList,key=lambda x:x[1])[:m])
score = [[ 'Angle', '0121701100106',99],[ 'Jack', '0121701100107',86],[ 'Tom', '0121701100109',65],[ 'Smith', '0121701100111',100],['Bob','0121701100115',77],['Lily','0121701100117',59]]
if n > len(score):
n = len(score)
print(sorted(score,key=lambda x:x[0])[:n])
print(sorted(score,key=lambda x:x[2])[:n])第5关 动物重量排序
ls = []
while True:
temp = input()
if temp == '':
break
ls.append(temp.split())
print(sorted(ls,key=lambda x:float(x[1][:-1])*1000 if x[1][-1] == 't' else float(x[1][:-2])))第6关 身份证号升位
n = input()
list1 = list(n)
Wi=[7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2]
if int(list1[6])+int(list1[7]) >= 5:
list1.insert(6,9)
list1.insert(6,1)
else:
list1.insert(6,0)
list1.insert(6,2)
s = 0
for i in range(17):
s += int(list1[i]) * int(Wi[i])
m = s % 11
list2 = [1,0,'X',9,8,7,6,5,4,3,2]
list1.append(list2[m])
for x in list1:
print(*str(x),end = '')第7关 完美立方数
N = int(input())
for a in range(2,N+1):
for b in range(2,a):
for c in range(b,a):
for d in range(c,a):
if a**3 == b**3+c**3+d**3:
print(f'Cube = {a},Triple = ({b},{c},{d})')第8关 约瑟夫环问题文章来源：https://www.toymoban.com/news/detail-855645.html
def Josephus(n,k):
ls_n=list(range(1,n+1))
while len(ls_n)>k-1:
ls_n=ls_n[k:]+ls_n[:k-1]
return ls_n
n,k=map(int,input().split())
if k>=2 and n>=k:
print(Josephus(n,k))
else:
print('Data Error!')第9关 统计英文文件中的单词数文章来源地址https://www.toymoban.com/news/detail-855645.html
def read_file(file):
"""接收文件名为参数，读取文件中的数据到字符串中，返回这个字符串"""
with open(file, 'r', encoding='utf-8') as text: # 创建文件对象
txt =text.read() # 读文件为字符串
return txt # 返回字符串
def word_list(txt):
"""接收字符串为参数，用空格替换字符串中所有标点符号，根据空格将字符串切分为列表
返回值为元素为单词的列表"""
for i in ",.!\'":
txt = txt.replace(i, ' ')
return txt.split()
def number_of_words(ls):
"""接收一个以单词为元素的列表为参数，返回列表中单词数量，返回值为整型"""
return len(ls)
if __name__ == '__main__':
filename = input() # 读入文件名
text = read_file('step10/'+filename) # 读取'step10/'文件夹中用户输入的文件名得到文件内容，存入text
words_list = word_list(text) # 处理text,得到单词的列表
words_counts = number_of_words(words_list) #统计单词列表word_list里的单词数
print(f'共有{words_counts}个单词')到了这里，关于python头歌-python第六章作业的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
