利用Python进行数据分析实验（一）

一、实验目的

使用Python解决简单问题

二、实验要求

自主编写并运行代码，按照模板要求撰写实验报告

三、实验步骤

本次实验共有5题：

有四个数字：1、2、3、4，能组成多少个互不相同且无重复数字的三位数？各是多少？四个数字是2、3、7、9呢？判断1000-2000之间有多少个素数，并输出所有素数.打印出所有的"四叶玫瑰数"，所谓"四叶玫瑰数"是指一个四位数，其各位数字四次方和等于该数本身。输入一行字符，分别统计出其中英文字母、空格、数字和其它字符输出的数并分别统计每一种类型的个数。打印九九乘法表。 四、实验结果 T1 T1-1 import itertools count = 0 for arr in itertools.permutations('1234', 3): # print(arr) print(int(arr[0]) * 100 + int(arr[1]) * 10 + int(arr[2])) count = count + 1 print('共有' + str(count) + '个组合') T1-2 import itertools count = 0 for arr in itertools.permutations('2379', 3): # print(arr) print(int(arr[0]) * 100 + int(arr[1]) * 10 + int(arr[2])) count = count + 1 print('共有' + str(count) + '个组合') T2 import math count = 0 res = [] def check(x): if x <= 1: return False for flag in range(2, int(math.sqrt(x) + 1)): if x % flag == 0: return False return True for i in range(1000, 2001): if check(i): res.append(i) count = count + 1 for i in range(0, len(res)): print(res[i], end=' ') if ((i + 1) % 10) == 0: print('\n') print('\n1000~2000有素数' + str(count) + '个') T3 def func(x): arr = str(x) if res(arr) == x: return True else: return False def res(arr): return pow(int(arr[0]), 4) + pow(int(arr[1]), 4) + pow(int(arr[2]), 4) + pow(int(arr[3]), 4) for i in range(1000, 10000): if func(i): print(i) T4 import re count_n = 0 count_s = 0 count_l = 0 count_o = 0 ''' string = 'Some people, when confronted with a problem, ' \ 'think “I know, I’ll use regular expressions.” ' \ 'Now they have two problems.' ''' string = input() number = re.finditer(r'\d+', string) for match in number: count_n = count_n + 1 print('数字的数量是' + str(count_n)) space = re.finditer(r'\s+', string) for match in space: count_s = count_s + 1 print('空白的数据是' + str(count_s)) letter = re.finditer(r'[a-zA-Z]', string) for match in letter: count_l = count_l + 1 print('字符的数量是' + str(count_l)) print('其他字符的数量是' + str(len(string) - count_s - count_l - count_n)) T5 import numpy as np import itertools count = 0 form = np.empty([81, 2], int) for arr in itertools.product('123456789', repeat=2): form[count][0] = arr[0] form[count][1] = arr[1] count = count + 1 print('打印99乘法表：') left = 0 count = 1 for i in range(0, 9): for j in range(left, left + count): string = str(form[j][0]) + '*' + str(form[j][1]) + '=' + str(int(form[j][0]) * int(form[j][1])) print('%-8s' % string, end='') print("\n") left = left + 9 count = count + 1 五、实验体会

Python标准库和第三库众多，功能强大。充分利用库函数来简化和加速代码、不重复造轮子能够显著简化代码提高编码效率