100+Python挑战性编程练习系列 -- day 3

Question 10

编写一个程序，接受一系列空格分隔的单词作为输入，并在删除所有重复的单词并按字母数字排序后打印这些单词。
假设向程序提供以下输入：
hello world and practice makes perfect and hello world again
然后，输出应为：
again and hello makes perfect practice world

方法1：

word = input().split()for i in word:if word.count(i) > 1:    #count function returns total repeatation of an element that is send as argumentword.remove(i)     # removes exactly one element per callword.sort()
print(" ".join(word))

方法2：

word = sorted(list(set(input().split())))              #  input string splits -> converting into set() to store unique#  element -> converting into list to be able to apply sort
print(" ".join(word))

Question 11

写一个程序，它接受一个由逗号分隔的4位二进制数序列作为输入，然后检查它们是否能被5整除。可被5整除的数字将以逗号分隔的顺序打印。
示例：
0100,0011,1010,1001
那么输出应该是：
1010

方法1：

def check(x):                       # converts binary to integer & returns zero if divisible by 5total,pw = 0,1reversed(x)for i in x:total+=pw * (ord(i) - 48)   # ord() function returns ASCII valuepw*=2return total % 5data = input().split(",")           # inputs taken here and splited in ',' position
lst = []for i in data:if check(i) == 0:               # if zero found it means divisible by zero and added to the listlst.append(i)print(",".join(lst))

方法2：

def check(x):                   # check function returns true if divisible by 5return int(x,2)%5 == 0      # int(x,b) takes x as string and b as base from which# it will be converted to decimal
data = input().split(',')data = list(filter(check,data)) # in filter(func,object) function, elements are picked from 'data' if found True by 'check' function
print(",".join(data))

方法3：

data = input().split(',')
data = [num for num in data if int(num, 2) % 5 == 0]
print(','.join(data))

Question 12

写一个程序，它将找到所有这样的数字之间的1000和3000（包括），使每个数字的数字是一个偶数。所获得的数字应以逗号分隔的顺序打印在一行上。

方法1：

lst = []for i in range(1000,3001):flag = 1for j in str(i):          # every integer number i is converted into stringif ord(j)%2 != 0:     # ord returns ASCII value and j is every digit of iflag = 0          # flag becomes zero if any odd digit foundif flag == 1:lst.append(str(i))    # i is stored in list as stringprint(",".join(lst))

方法2：

def check(element):return all(ord(i)%2 == 0 for i in element)  # all returns True if all digits i is even in elementlst = [str(i) for i in range(1000,3001)]        # creates list of all given numbers with string data type
lst = list(filter(check,lst))                   # filter removes element from list if check condition fails
print(",".join(lst))

方法3：（一行）

print(','.join([str(num) for num in range(1000, 3001) if all(map(lambda num: int(num) % 2 == 0, str(num)))]))

Question 13

写一个程序，接受一个句子并计算字母和数字的数量。
假设向程序提供以下输入：
hello world! 123
然后，输出应为：
LETTERS 10
DIGITS 3

方法1：

word = input()
letter,digit = 0,0for i in word:if ('a'<=i and i<='z') or ('A'<=i and i<='Z'):letter+=1if '0'<=i and i<='9':digit+=1print("LETTERS {0}\nDIGITS {1}".format(letter,digit))

方法2：

word = input()
letter, digit = 0,0for i in word:if i.isalpha(): # returns True if alphabetletter += 1elif i.isnumeric(): # returns True if numericdigit += 1
print(f"LETTERS {letter}\n{digits}") # two different types of formating method is shown in both solution

方法3：

#using reduce for to count
from functools import reducedef count_letters_digits(counters,char_to_check):counters[0] += char_to_check.isalpha()counters[1] += char_to_check.isnumeric()return countersprint('LETTERS {0}\nDIGITS {1}'.format(*reduce(count_letters_digits,input(),[0,0])))