Question

这是一个多肉的问题，但对于使用list和tuples的其他人来说可能会有用。我需要获取列表CALC_RECORD，然后从列表中拉出第三个元素。输出目前看起来像这样：[('calculation', 1467938304.345363, 1.2636651992797852)]。元组中的第三个元素是玩家回答问题所花费的时间。每次回答问题时，都会在列表中创建一个新元组。所以它看起来像这样：[('calculation', 1467938302.2010334, None), ('calculation', 1467938302.8568625, None), ('calculation', 1467938304.345363, 1.2636651992797852)]。我需要从每个元组中提取第三个元素，并使用这些元素创建一个列表。然后该列表将用作图表中的y轴（x轴是游戏的总次数。我尝试使用带有过滤功能的lambda，但我无处可去。有什么想法吗？

注：

以下程序的完整版本中有更多游戏/功能，但它们已被移除以节省空间，因为如果可以回答问题，则可以对它们使用相同的答案。因此，将它们放在这里将是多余的。

编辑：我试着这样做。

#EDIT: filtered_list = list(filter(lambda x: x[0] =='calculation' and x[2] != None, CALC_RECORD))

import random
from random import randint
import time
import math
import matplotlib.pyplot as plt

# Number of problems for each practice/real round
practice_round = 0
real_round = 3

main_record = []
CALC_RECORD = []

# (1) Calculation Game ---------------------------------------------------------

def calculation():
    response_time = None
    # Determine the min and max calculation values
    min_calculation_value = 1
    max_calculation_value = 10
    # Generate the problems
    print('\nSolve the following problem:')
    a = random.randint(min_calculation_value, max_calculation_value)
    b = random.randint(min_calculation_value, max_calculation_value)
    problem_type = random.randint(1,2)
    if problem_type == 1:
        answer = a * b
        print(a, '*', b)
    elif problem_type == 2:
        answer = a % b
        print(a, '%', b)
    # Get the user's answer determine what to do if correct
    start_time = time.time()
    user_answer = input('\nEnter your answer: ')
    end_time = time.time()
    if user_answer == str(answer):
        response_time = end_time - start_time
        print('You are correct!')
    elif user_answer != str(answer):
        print('You are incorrect.')
    # Return game id, start time, and response time
    return("calculation", start_time, response_time)

def calculation_game():
    record = []
    # Generate two problems for a practice round
    print("\nLet's begin with 2 practice problems.")
    for i in range (practice_round):
        print('\nPractice Problem', i + 1, 'of', practice_round)
        calculation()
    # Generate 10 problems for a real, recorded round
    print("\nNow let's try it for real this time.")
    for i in range (real_round):
        print('\nProblem', i + 1, 'of', real_round)
        # Append records for each iteration
        record.append(calculation())
    main_record.extend(record)
    CALC_RECORD.extend(record)
    return record

# (5) Display Data -------------------------------------------------------------
def display_data():
    print (CALC_RECORD)
#This function is currently just being used to view the output of calc record

----------------------------------------------------------------

def quit_game():
    print('\nThank you for playing!')

# Main Menu --------------------------------------------------------------------

def menu():
    print("\nEnter 1 to play 'Calculation'")
    print("Enter 2 to play 'Binary Reader'")
    print("Enter 3 to play 'Trifacto'")
    print("Enter 4 to view your statistics")
    print("Enter 5 to display data")
    print("Enter 6 to save your progress")
    print("Enter 7 to load data")
    print("Enter 8 to quit the game")

def main_menu():

    print('Welcome--Let's Play!')
    main_record = []
    user_input = ''
    while user_input != '8':
        menu()
        user_input = input('\nWhat would you like to do? ')
        if user_input == '1':
            calculation_game()
        if user_input == '2':
            binary_reader_game()
        if user_input == '3':
            trifacto_game()
        if user_input == '4':
            display_statistics()
        if user_input == '5':
            display_data()
        if user_input == '8':
            quit_game()

main_menu()

Answer 1

如果你有一个元组列表，并且想要将每个元组的第三个元素拉出到一个新列表中，那么以下内容应该可以工作：

new_list = [t[2] for t in old_list]

或者您可以使用map和lambda：

new_list = list(map(lambda t: t[2], old_list))

In [11]: old_list = [('calculation', 1467938302.2010334, None), ('calculation', 1467938302.8568625, None), ('calculation', 1467938304.345363, 1.2636651992797852)]

In [12]: old_list
Out[12]: 
[('calculation', 1467938302.2010334, None),
 ('calculation', 1467938302.8568625, None),
 ('calculation', 1467938304.345363, 1.2636651992797852)]

In [13]: [t[2] for t in old_list]
Out[13]: [None, None, 1.2636651992797852]

In [14]: list(map(lambda t:t[2], old_list))
Out[14]: [None, None, 1.2636651992797852]

编辑：根据您添加初始尝试的代码，您可以使用列表理解轻松映射和过滤值，如下所示：

In [15]: [t[2] for t in old_list if t[0] =='calculation' and t[2] != None]
Out[15]: [1.2636651992797852]

需要帮助创建和绘制列表

1 个答案: