From bdcadf146a2c23f76e88295f67ad29453c2bc272 Mon Sep 17 00:00:00 2001 From: XueqiMa <13363241795@163.com> Date: Sun, 10 Apr 2022 11:17:17 +0000 Subject: [PATCH] 202122011079 --- 202122011079.ipynb | 178 +++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 178 insertions(+) create mode 100644 202122011079.ipynb diff --git a/202122011079.ipynb b/202122011079.ipynb new file mode 100644 index 0000000..3a30c31 --- /dev/null +++ b/202122011079.ipynb @@ -0,0 +1,178 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "id": "82f0792c", + "metadata": {}, + "source": [ + "# 题目一\n", + "有一个这样的DNA核酸序列\n", + "\n", + "“AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC”\n", + "\n", + "请把这个核酸序列存入一个list，并数一数A、G、C、T各有多少个。" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "id": "acf452ea", + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "此DNA核酸序列中有20个A, 17个G, 12个C, 21个T\n" + ] + } + ], + "source": [ + "alist = list('AGCTTTTCATTCTGACTGCAACGGGCAATATGTCTCTGTGTGGATTAAAAAAAGAGTGTCTGATAGCAGC')\n", + "num_A = alist.count('A')\n", + "num_G = alist.count('G')\n", + "num_C = alist.count('C')\n", + "num_T = alist.count('T')\n", + "print(\"此DNA核酸序列中有%d个A, %d个G, %d个C, %d个T\"%(num_A,num_G,num_C,num_T))" + ] + }, + { + "cell_type": "markdown", + "id": "34ada2bd", + "metadata": {}, + "source": [ + "# 题目二\n", + "一个花样滑冰运动员表演后，裁判给表演内容进行评分，分数从0.25分到10分，每次增加值为0.25分。\n", + "\n", + "试生成一个元组，把可能的得分存入元组，并遍历元组的每一项，打印“一个运动员可能得_____分”。" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "id": "4df27377", + "metadata": { + "scrolled": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "一个运动员可能得0.25分\n", + "一个运动员可能得0.50分\n", + "一个运动员可能得0.75分\n", + "一个运动员可能得1.00分\n", + "一个运动员可能得1.25分\n", + "一个运动员可能得1.50分\n", + "一个运动员可能得1.75分\n", + "一个运动员可能得2.00分\n", + "一个运动员可能得2.25分\n", + "一个运动员可能得2.50分\n", + "一个运动员可能得2.75分\n", + "一个运动员可能得3.00分\n", + "一个运动员可能得3.25分\n", + "一个运动员可能得3.50分\n", + "一个运动员可能得3.75分\n", + "一个运动员可能得4.00分\n", + "一个运动员可能得4.25分\n", + "一个运动员可能得4.50分\n", + "一个运动员可能得4.75分\n", + "一个运动员可能得5.00分\n", + "一个运动员可能得5.25分\n", + "一个运动员可能得5.50分\n", + "一个运动员可能得5.75分\n", + "一个运动员可能得6.00分\n", + "一个运动员可能得6.25分\n", + "一个运动员可能得6.50分\n", + "一个运动员可能得6.75分\n", + "一个运动员可能得7.00分\n", + "一个运动员可能得7.25分\n", + "一个运动员可能得7.50分\n", + "一个运动员可能得7.75分\n", + "一个运动员可能得8.00分\n", + "一个运动员可能得8.25分\n", + "一个运动员可能得8.50分\n", + "一个运动员可能得8.75分\n", + "一个运动员可能得9.00分\n", + "一个运动员可能得9.25分\n", + "一个运动员可能得9.50分\n", + "一个运动员可能得9.75分\n", + "一个运动员可能得10.00分\n" + ] + } + ], + "source": [ + "import numpy as np\n", + "score = tuple(np.arange(0.25,10.25,0.25))\n", + "for i in range(len(score)):\n", + " print(\"一个运动员可能得%.2f分\"%score[i])" + ] + }, + { + "cell_type": "markdown", + "id": "df88a6cd", + "metadata": {}, + "source": [ + "# 题目三\n", + "创建一个字典，列出你所了解的地域美食，比如{'肠粉':{'城市':'广州','原料':'米'}}。当然，你可以做的更丰富一些。 最后遍历你熟悉的美食，打印出，类似如下的句子：“肠粉是广州的一种美食，它的主要原料是米”。" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "id": "8fef1067", + "metadata": {}, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "肠粉是广州的一种美食，它的主要原料是米\n", + "火锅是成都的一种美食，它的主要原料是蔬菜\n", + "烧烤是沈阳的一种美食，它的主要原料是羊肉\n" + ] + } + ], + "source": [ + "a = {}\n", + "a['肠粉'] = {'城市':'广州','原料':'米'}\n", + "a['火锅'] = {'城市':'成都','原料':'蔬菜'}\n", + "a['烧烤'] = {'城市':'沈阳','原料':'羊肉'}\n", + "for a,b in a.items():\n", + " city = b['城市']\n", + " c = b['原料']\n", + " print(\"%s是%s的一种美食，它的主要原料是%s\"%(a,city,c))" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "id": "996c7f49", + "metadata": {}, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.8.8" + } + }, + "nbformat": 4, + "nbformat_minor": 5 +} -- Gitee