有以下两列表list_1=["a","b","e","f","d","c","g"]；list_2=["c","d","w","z","b","f"]找出两列表共有的元素（提示：可以采用集合的方式），并将其添加到一个新列表中

修改下面代码，另画一张可视化图展示出t_sne里面的数据每15行数据个用一种颜色画出。 import pandas as pd from sklearn import cluster from sklearn import metrics import matplotlib.pyplot as plt from sklearn.manifold import TSNE from sklearn.decomposition import PCA def k_means(data_set, output_file, png_file, t_labels, score_file, set_name): model = cluster.KMeans(n_clusters=7, max_iter=1000, init="k-means++") model.fit(data_set) # print(list(model.labels_)) p_labels = list(model.labels_) r = pd.concat([data_set, pd.Series(model.labels_, index=data_set.index)], axis=1) r.columns = list(data_set.columns) + [u'聚类类别'] print(r) # r.to_excel(output_file) with open(score_file, "a") as sf: sf.write("By k-means, the f-m_score of " + set_name + " is: " + str(metrics.fowlkes_mallows_score(t_labels, p_labels))+"\n") sf.write("By k-means, the rand_score of " + set_name + " is: " + str(metrics.adjusted_rand_score(t_labels, p_labels))+"\n") '''pca = PCA(n_components=2) pca.fit(data_set) pca_result = pca.transform(data_set) t_sne = pd.DataFrame(pca_result, index=data_set.index)''' t_sne = TSNE() t_sne.fit(data_set) t_sne = pd.DataFrame(t_sne.embedding_, index=data_set.index) plt.rcParams['font.sans-serif'] = ['SimHei'] plt.rcParams['axes.unicode_minus'] = False dd = t_sne[r[u'聚类类别'] == 0] plt.plot(dd[0], dd[1], 'r.') dd = t_sne[r[u'聚类类别'] == 1] plt.plot(dd[0], dd[1], 'go') dd = t_sne[r[u'聚类类别'] == 2] plt.plot(dd[0], dd[1], 'b') dd = t_sne[r[u'聚类类别'] == 3] plt.plot(dd[0], dd[1], 'o') dd = t_sne[r[u'聚类类别'] == 4] plt.plot(dd[0], dd[1], 'm.') dd = t_sne[r[u'聚类类别'] == 5] plt.plot(dd[0], dd[1], 'co') dd = t_sne[r[u'聚类类别'] == 6] plt.plot(dd[0], dd[1], 'y') plt.savefig(png_file) plt.clf() '''plt.scatter(data_set.iloc[:, 0], data_set.iloc[:, 1], c=model.labels_) plt.savefig(png_file) plt.clf()''' frog_data = pd.read_csv("D:/PyCharmPython/pythonProject/mfcc3.csv") tLabel = [] for family in frog_data['name']: if family == "A": tLabel.append(0) elif family == "B": tLabel.append(1) elif family == "C": tLabel.append(2) elif family == "D": tLabel.append(3) elif family == "E": tLabel.append(4) elif family == "F": tLabel.append(5) elif family == "G": tLabel.append(6) scoreFile = "D:/PyCharmPython/pythonProject/scoreOfClustering.txt" first_set = frog_data.iloc[:, 1:1327] k_means(first_set, "D:/PyCharmPython/pythonProject/kMeansSet_1.xlsx", "D:/PyCharmPython/pythonProject/kMeansSet_2.png", tLabel, scoreFile, "Set_1")

k_means(first_set, "D:/PyCharmPython/pythonProject/kMeansSet_1.xlsx", "D:/PyCharmPython/pythonProject/kMeansSet_2.png", tLabel, scoreFile, "Set_1") 这段代码对原始代码进行了修改，使用了循环来绘制 ...

file = '新建文本文档.txt' with open(file,encoding='utf-8') as f: lines = f.readlines() print('lines为：',lines) split_list = [] for s in lines: s = s.strip() # 删除字符串两端的空格和换行符 #print('s为',s) split_list.append(s.split(':', 1)) # 用第一个冒号分割字符串，最多分割一次 first_list = [s[0] for s in split_list] second_list = [s[1] for s in split_list] new_second_list=[] print(first_list) print(second_list) for i in second_list: i = i.strip() # 删除字符串两端的空格和换行符 new_second_list.append(i) print(new_second_list) #转化为列表 dict = dict(zip(first_list,new_second_list)) print(dict) a = [] # SrcDatabase-来源库 b = [] # Title-题名 c = [] # Author-作者 d = [] # Organ-单位 e = [] # Source-文献来源 f = [] # PubTime-发表时间 g = [] # Volume-卷 h = [] # Period-期 for key,value in dict.items(): if 'SrcDatabase-来源库' in dict[key]: a.append(dict[value]) elif 'Title-题名' in dict[key]: b.append(dict[value]) elif 'Author-作者' in dict[key]: c.append(dict[value]) elif 'Organ-单位' in dict[key]: d.append(dict[value]) elif 'Source-文献来源' in dict[key]: e.append(dict[value]) elif 'PubTime-发表时间' in dict[key]: f.append(dict[value]) elif 'Volume-卷' in dict[key]: g.append(dict[value]) else: h.append(dict[value]) print('SrcDatabase-来源库:',a) print('Title-题名:',b) print('Author-作者:',c) print('Organ-单位:',d) print('Source-文献来源:',e) print('PubTime-发表时间',f) print('Volume-卷:',g) print('Period-期:',h) df = pd.DataFrame({"SrcDatabase-来源库":a,"Title-题名": b,"Author-作者":c,"Organ-单位": d,"Source-文献来源":e,"PubTime-发表时间": f,"Volume-卷":g,"Period-期": h}) df.to_excel("处理后的文档.xlsx", index=False)有问题吗，这个代码，如有，怎么解决

这段代码的作用是打开一个名为'新建文本文档.txt'的文件，并以utf-8的编码方式读取...接着对lines中每一个字符串进行去除空白符的操作（strip()函数），并把去除空白符后的字符串存储在一个新的列表变量split_list中。

def guess_key3(cipher_text, word1, word2, word3): #变了点 letter_frequency = get_letter_frequency(cipher_text.lower()) excluded_letters = [letter for letter in letter_frequency.keys() if letter_frequency[letter] == 0] sorted_letters = sorted([letter for letter in letter_frequency.keys() if letter_frequency[letter] > 0], key=lambda x: letter_frequency[x], reverse=True) print("Excluded letters:", excluded_letters) print() f1 = ['a', 'i', 'r'] f2 = ['t', 'o', 'n'] f3 = ['s', 'l', 'c'] f4 = ['u', 'p', 'm', 'd', 'h'] f5 = ['g', 'b', 'y', 'f'] f6 = ['v', 'w','k'] f7 = ['x', 'z', 'q', 'j'] mf = [f6, f5, f4, f3, f2, f1] key = {sorted_letters[0]: 'e'} most_common_letters_m = [sorted_letters[19:22],sorted_letters[15:19] ,sorted_letters[10:15] ,sorted_letters[7:10] ,sorted_letters[4:7], sorted_letters[1:4]] c1 = 0.05 for i1 in range(5): lk1=len(key) key1 = check3(cipher_text, word1, word2, word3, most_common_letters_m, sorted_letters, mf, f7, lk1, key, c1, i1) key.update(key1) del mf[-1] del most_common_letters_m[-1] print(key) print() c1+=0.1 return key def check3(cipher_text, word1, word2, word3, most_common_letters_m, sorted_letters, mf, f7, lk1, key, c1, i1): mp = [[j for j in range(len(mf[i]))] for i in range(len(mf))] row_permutations = [itertools.permutations(row) for row in mp] matrix_permutations = itertools.product(*row_permutations) for permutation in matrix_permutations: for i in range(len(mp)): for j in range(len(mp[i])): key[most_common_letters_m[i][permutation[i][j]]] = mf[i][j] if len(key) < len(sorted_letters): for i, val in enumerate(f7): key[sorted_letters[i]] = val decrypted_text = decrypt(cipher_text, key) k1 = is_plaintext3(decrypted_text, word1, word2 ,word3) #k2 = k1 if k1 > k2 else k2 if( k1 > c1): key1 = dict(list(key)[lk1:len(mp[i1])+lk1]) return key1 def is_plaintext3(text, word1, word2 ,word3): words_found = 0 for word in text.split(): if word.lower() in word1: words_found += 10 if word.lower() in word2: words_found += 3 if word.lower() in word3: words_found += 1 return (words_found / len(text.split())) def decrypt(cipher_text, key): mapping_dict = str.maketrans(key) return cipher_text.translate(mapping_dict)有什么问题

1. 在 check3 函数中，如果没有找到有效的解密密钥，则没有返回值，因此当函数结束时会返回 None。这可能会导致 guess_key3 函数出现异常或错误，因为它不会检查 check3 的返回值是否为 None。 2. 在 check3 函数中...

list1=0,1,2,3,4,5,6,7… list2=A,B,C,D,E,F,G... c# form窗体，我想生成一个list3=0-A,1-B,2-C,3-D怎么实现

List<char> list2 = new List() { 'A', 'B', 'C', 'D', 'E', 'F', 'G' }; // 使用Zip()方法结合两个列表 List<string> list3 = list1.Zip(list2, (x, y) => $"{x}-{y}") .ToList(); 这个例子中，Zip()方法...

def read_device_excel( ): ip_list = [] try: filepath = askopenfilename(filetypes=[("Excel files", "*.xlsx")]) wb1 = load_workbook(filepath) ws1 = wb1.get_sheet_by_name("Sheet1") for cow_num in range(2, ws1.max_row+1): ipaddr = ws1["B"+str(cow_num)].value ntos = ws1["C"+str(cow_num)].value name = ws1["D"+str(cow_num)].value pswd = ws1["E"+str(cow_num)].value epswd = ws1["F"+str(cow_num)].value inscmd = ws1["G"+str(cow_num)].value ip_list.append((ipaddr, ntos, name, pswd, epswd, inscmd)) # print (ip_list) return ip_list except Exception as e: print("读取Excel文件出错：", e) return [] root = Tk() button = Button(root, text="选择文件", command=read_device_excel) button.pack() root.mainloop() def get_config(ipaddr,ntos,name,pswd,epswd,inscmd): session = ConnectHandler( device_type=ntos, ip=ipaddr, username=name, password=pswd, secret=epswd, port = 22, verbose = True, banner_timeout=300 )上传文件后并没有继续运行程序

以下是修改后的代码： def get_config(ipaddr, ntos, name, pswd, epswd, inscmd): try: session = ConnectHandler( device_type=ntos, ip=ipaddr, username=name, password=pswd, secret=epswd, port=...

def futurebalance(): while True: a = input("请输入您要计算资产的时间(年):") if a == "q": break elif a == "0": print("输入有误,请重新输入") elif a.isdigit(): b = user_list[account]["贷款年限"] if int(a) > b or int(a) == b: c = interest_list["贷款利率"] d = user_list[account]["贷款"] e = d+(bcd) f = interest_list["存款利率"] g = user_list[account]["余额"] h = (g*(1+f)**int(a))-e print("您" + a + "年后的资产为:" + str(h)) else: c2 = user_list[account]["余额"] d2 = interest_list["存款利率"] e2 = c2 * ((1 + d2) ** int(a)) print("您" + a + "年后的资产为:" + str(e2))

如果用户输入“0”，函数将提示输入有误，并要求用户重新输入。如果用户输入的是数字，函数将计算未来资产。如果输入的时间大于贷款年限或等于贷款年限，函数将计算用户未来资产时考虑贷款的影响。否则，函数将只...

from itertools import combinations # 定义数组 A = [1,3,4,5,8,10,12,13,14,16,17,20,21,25,26,28] B = [1,3,4,7,11,12,15,16,19,20,22,24,28,31,32] C = [2,7,10,13,15,16,19,20,22,25,26,28,29,31,33] D = [1,2,3,5,8,10,14,15,17,25,26,29,33] E = [3,5,8,11,12,15,19,21,24,26,30,31,33] F = [3,6,8,10,12,14,15,17,20,21,29,31] G = [3,4,7,12,14,16,19,21,25,28,29,31] H = [4,6,10,11,12,18,19,20,22,25,26,31] # 合并所有数组 arrays = [A,B,C,D,E,F,G,H] # 生成所有可能的6位数顺序组合 combinations_list = list(combinations([num for array in arrays for num in array], 6)) # 去除重复的组合 unique_combinations = list(set(combinations_list)) # 将组合转化为字符串 unique_combinations = ["".join(map(str, combination)) for combination in unique_combinations] print(unique_combinations)

这段代码看起来是正确的，它将打印出所有不重复的6位数顺序组合。它使用了 itertools 模块的 combinations 函数来生成所有可能的组合，并通过转换为字符串的方式去除了重复的...如果你有任何其他问题，请随时提问！

class CIFAR10(Dataset): """CIFAR10 <https://www.cs.toronto.edu/~kriz/cifar.html>_ Dataset. Args: root (string): Root directory of dataset where directory cifar-10-batches-py exists or will be saved to if download is set to True. train (bool, optional): If True, creates dataset from training set, otherwise creates from test set. transform (callable, optional): A function/transform that takes in an PIL image and returns a transformed version. E.g, transforms.RandomCrop download (bool, optional): If true, downloads the dataset from the internet and puts it in root directory. If dataset is already downloaded, it is not downloaded again. """ base_folder = 'cifar-10-batches-py' url = "https://www.cs.toronto.edu/~kriz/cifar-10-python.tar.gz" filename = "cifar-10-python.tar.gz" tgz_md5 = 'c58f30108f718f92721af3b95e74349a' train_list = [ ['data_batch_1', 'c99cafc152244af753f735de768cd75f'], ['data_batch_2', 'd4bba439e000b95fd0a9bffe97cbabec'], ['data_batch_3', '54ebc095f3ab1f0389bbae665268c751'], ['data_batch_4', '634d18415352ddfa80567beed471001a'], ['data_batch_5', '482c414d41f54cd18b22e5b47cb7c3cb'], ['test_batch', '40351d587109b95175f43aff81a1287e'], ] test_list = [ ['test_batch', '40351d587109b95175f43aff81a1287e'], ] meta = { 'filename': 'batches.meta', 'key': 'label_names', 'md5': '5ff9c542aee3614f3951f8cda6e48888', }

在类的定义中，有以下几个重要的属性和方法： - base_folder：CIFAR-10数据集的基础文件夹名称。 - url：CIFAR-10数据集的下载链接。 - filename：CIFAR-10数据集压缩文件的名称。 - tgz_md5：CIFAR-10数据...

检测鼠标事件 def mouse_event(self, event, x, y, flags, param): if event == cv2.EVENT_LBUTTONUP and x > 550 and y < 50: def open_login_window(my_window, on_entry_click): loginwindow = LoginWindow(on_entry_click) loginwindow.transient(my_window) loginwindow.wait_visibility() loginwindow.grab_set() def quit_window(my_window): # self.camera_process.terminate() my_window.destroy() # 虚拟键盘 def on_entry_click(self, event, entry): if self.keyboard_window: self.keyboard_window.destroy() keyboard_window = tk.Toplevel(self) keyboard_window.title("虚拟键盘") keyboard_window.geometry("610x140") keyboard_window.resizable(False, False) button_list = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0', '<-', 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p', 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l', 'z', 'x', 'c', 'v', 'b', 'n', 'm'] row = 0 col = 0 for button_text in button_list: button = tk.Button(keyboard_window, text=button_text, width=3) if button_text != '<-': button.config(command=lambda char=button_text: entry.insert(tk.END, char)) else: button.config( command=lambda char=button_text: entry.delete(len(entry.get()) - 1, tk.END)) button.grid(row=row, column=col) col += 1 if col > 10: row += 1 col = 0 keyboard_window.deiconify() self.keyboard_window = keyboard_window # 登录界面 my_window = tk.Tk() my_window.title("登录") my_window.geometry("300x200") # 计算窗口位置，让其出现在屏幕中间 screen_width = my_window.winfo_screenwidth() screen_height = my_window.winfo_screenheight() x = (screen_width - 300) // 2 y = (screen_height - 200) // 2 my_window.geometry("+{}+{}".format(x, y)) my_window.wm_attributes("-topmost", True) login_button = tk.Button(my_window, text="登录", font=('Arial', 12), width=10, height=1, command=lambda: open_login_window(my_window, on_entry_click)) login_button.pack(side='left', expand=True) exitbutton = tk.Button(my_window, text="退出", font=('Arial', 12), width=10, height=1, command=lambda: [quit_window(my_window)]) exitbutton.pack(side='left', expand=True) my_window.mainloop() if event == cv2.EVENT_LBUTTONUP and x < 50 and y > 1000: cv2.destroyAllWindows() 在此基础上请实现让tk界面不会出现重影用中文回答

import cv2 import threading import tkinter as tk class GUI: def __init__(self): self.root = tk.Tk() self.label = tk.Label(self.root, text='Hello, world!') self.label.pack() self.thread = ...

library(xlsx) setwd("D:/R") # 读取数据 data1 <- read.xlsx("试验.xlsx",sheetName = "man") data2 <- read.xlsx("试验.xlsx",sheetName = "weman") data3 <- read.xlsx("试验.xlsx",sheetName = "all") B <- data1$男孩 F <- data3$父亲 G <- data2$女孩 M <- data3$母亲 # t检验 t1=(mean(B)-mean(F))/(sd(B)/sqrt(length(B))) t1 p1 <- pt(t1,length(B)-1) p1 if (p1>0.05){ print("男生高于父亲") } else{ print("男生不高于父亲") } #检验结果为t=-5.123508，显著性P值=1.488898e-06>0.05，接受原假设，认为男生不高于父亲 t2=(mean(G)-mean(M))/(sd(G)/sqrt(length(G))) t2 p2 <- pt(t2,length(G)-1) p2 if (p2>0.05){ print("女生高于母亲") } else{ print("女生不高于母亲") } #检验结果为t=3.994843，显著性P值=0.9999377>0.05，接受原假设，认为女生高于母亲 # 计算结果 results <- list() # 中亲 - 子女 x <- data3$学生 y <- data3$中亲 lm.model <- lm(x ~ y, data = data3) model_summary <- summary(lm.model) model_coeffs <- coef(lm.model) standard_error <- model_summary$sigma CI <- confint(lm.model) cor_coef <- cor(x, y) p_value <- model_summary$coefficients[2,4] results[["中亲-子女"]] <- data.frame(coefficients = model_coeffs, std_error = standard_error, conf_int = CI, cor_coef = cor_coef, p_value = p_value)结果分析

这是一段 R 代码，用于进行 t 检验和线性回归分析。代码中分别读取了三个 Excel 表格的数据，并对其中的数据进行了处理和分析，包括了男孩身高与父亲身高的 t 检验以及女孩身高与母亲身高的 t 检验，以及中亲身高与...

解释以下代码。def get_goods_list(page_size, page_no, param): db = dbUtil() param = param.replace("\\", "") count_sql = "select count() from goods where " + param count_res = db.query(count_sql)[0][0] start = page_size (page_no - 1) start = 0 if start < 0 else start sql = "select * from goods where" + param + "order by id desc limit " + str(start) + "," + str(page_size) res = db.query(sql) data_page = [] if count_res % page_size == 0: max_page = int(count_res / page_size) else: max_page = int(count_res / page_size) + 1 if max_page <= 5: page_list = [i for i in range(1, max_page + 1, 1)] elif page_no + 2 > max_page: page_list = [i for i in range(max_page - 5, max_page + 1, 1)] elif page_no - 2 < 1: page_list = [i for i in range(1, 6, 1)] else: page_list = [i for i in range(page_no - 2, page_no + 3, 1)] for a, b, c, d, e, f, g, h in res: item = [a, b, c, d, e, f, g, h] data_page.append(item) db.close() return data_page, count_res, page_list, max_page

这段代码是一个函数，函数名为get_goods_list，它有三个参数：page_size（每页显示的数据条数），page_no（当前页码），param（查询条件）。该函数会从数据库中查询符合条件的商品列表数据，并返回一个包含数据列表...

2、有以下两列表list_1=["a","b","e","f","d","c","g"]；list_2=["c","d","w","z","b","f"]找出两列表共有的元素（提示：可以采用集合的方式），并将其添加到一个新列表中

list_1 = ["a", "b", "e", "f", "d", "c", "g"] list_2 ["c", "d", "w", "z", "b", "f"] # 将两个列表转换为集合 set_1 = set(list_1) set_2 = set(list_2) # 找出两个集合的交集 common_elements = list(set_1....

将二维数组 my_list = [['a b c'], ['b d e'], ['c f g']] 变成 my_list = [[a b c],[b d e],[c f g]]

要将二维数组 my_list = [['a b c'], ['b d e'], ['c f g']] 变成 my_list = [['a', 'b', 'c'], ['b', 'd', 'e'], ['c', 'f', 'g']]，可以使用以下代码： python my_list = [['a b c'], ['b d e'], ['c f g']] ...

test.py homework.py import random Ochoice_list = ['0', '1', '2', '3', '4','5', '6', '7', '8', '9' 'a','b', 'c', 'd', 'e','f','g', 'h', 'i','g'] Dif name == "main": str_len = random.randint(1, 1000) final_str =! for i in range(str_len): final str += random.choice(choice_list) print(str_len) print(final_str)

choice_list = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i'] if __name__ == "__main__": str_len = random.randint(1, 1000) final_str = "" for i in...

我的第一个C#小程序之简单音乐播放器1731655933.html

有以下两列表list_1=["a","b","e","f","d","c","g"]；list_2=["c","d","w","z","b","f"]找出两列表共有的元素（提示：可以采用集合的方式），并将其添加到一个新列表中

相关推荐

在list集合中输入元素，去除重复的元素并输出

Python列表list操作相关知识小结

华为leetcode-LeetCode-Huawei_OJ:LeetCode&Huawei_OJ测试用例

list1=0,1,2,3,4,5,6,7… list2=A,B,C,D,E,F,G... c# form窗体，我想生成一个list3=0-A,1-B,2-C,3-D怎么实现

2、有以下两列表list_1=["a","b","e","f","d","c","g"]；list_2=["c","d","w","z","b","f"]找出两列表共有的元素（提示：可以采用集合的方式），并将其添加到一个新列表中

将二维数组 my_list = [['a b c'], ['b d e'], ['c f g']] 变成 my_list = [[a b c],[b d e],[c f g]]

我的第一个C#小程序之简单音乐播放器1731655933.html

最新推荐

我的第一个C#小程序之简单音乐播放器1731655933.html

练习springboot1 项目 模拟高并发秒杀，实现基本的登录、查看商品列表、秒杀、下单等功能，简单实现了系统缓存、降级和限流

MATLAB实现小波阈值去噪：Visushrink硬软算法对比

管理建模和仿真的文件

【交互特征的影响】：分类问题中的深入探讨，如何正确应用交互特征

c语言从链式队列 中获取头部元素并返回其状态的函数怎么写

易语言实现画板图像缩放功能教程

"互动学习：行动中的多样性与论文攻读经历"

【交互特征：优化与调试的艺术】：实战技巧，提升回归模型与分类模型的性能

用IDEA写一个高速收费系统框架附带代码

练习springboot1 项目模拟高并发秒杀，实现基本的登录、查看商品列表、秒杀、下单等功能，简单实现了系统缓存、降级和限流

c语言从链式队列中获取头部元素并返回其状态的函数怎么写