Question

def open_marks():
    N = (int(input('how many students? '))* 5)
    students = []
    for line in open('marks.txt').readlines():
       datafile = (line.strip().split('\t')[0].split(','))  
       for n in datafile:
          students.append(int(n))
    students=students[:N]
    return students
def open_marks1():
    students = open_marks()
    students1=students[0::5]#set to return only the first(lowest) marks drawn
    return students1
def open_marks2():
    students = open_marks()
    students2=students[1::5]#set to return only the second marks drawn
    return students2
def open_marks3():
    students = open_marks()
    students3=students[2::5]#set to return only the third marks drawn
    return students3
def open_marks4():
    students = open_marks()
    students4=students[3::5]#set to return only the fourth marks drawn
    return students4
def open_marks5():
    students = open_marks()
    students5=students[4::5]#set to return only the fifth(highest) marks drawn
    return students5


def count_ranges_one():

    students1 = open_marks1()
    print('first number: ',students1)
    range_counts1 = [0] * 12
    for num in students1[:]:#change number to select number of draws
        which_range=int(num//5)
        range_counts1[which_range] = range_counts1[which_range] + 1
    return range_counts1
def count_ranges_two():
    students2 = open_marks2()
    print('second number: ',students2)
    range_counts2 = [0] * 12
    for num in students2[:]:#change number to select number of draws
        which_range=int(num//5)
        range_counts2[which_range] = range_counts2[which_range] + 1
    return range_counts2
def count_ranges_three():
    students3 = open_marks3()
    print('third number: ',students3)
    range_counts3 = [0] * 12
    for num in students3[:]:#change number to select number of draws
        which_range=int(num//5)
        range_counts3[which_range] = range_counts3[which_range] + 1
    return range_counts3
def count_ranges_four():
    students4 = open_marks4()
    print('fourth number: ',students4)
    range_counts4 = [0] * 12
    for num in students4[:]:#change number to select number of draws
        which_range=int(num//5)
        range_counts4[which_range] = range_counts4[which_range] + 1
    return range_counts4
def count_ranges_five():
    students5 = open_marks5()
    print('fifth number: ',students5)
    range_counts5 = [0] * 12
    for num in students5[:]:#change number to select number of draws
        which_range=int(num//5)
        range_counts5[which_range] = range_counts5[which_range] + 1
    return range_counts5

我有一个带有学生标记的文本文件格式： 7,5,10,25,32 9,15,25,39,18 等等上面根据位置（1,2,3,4或5）对标记进行分组，然后有一个例程来'写'直方图'来显示标记的分布。我写的非常笨拙和重复，但我无法弄清楚循环函数来提供每个问题分组为标记的数据，然后是直方图的每个范围的数字。有人可以帮我解决这个问题。

Answer 1

您知道您可以将参数传递给您的函数吗？五个open_marks[1-5]函数可以这样概括：

def open_marks_for_student(n):
    students = open_marks()
    return students[n-1::5]

如果您现在通过，例如3，对于该函数，您将获得与旧的open_marks3函数相同的结果：

open_marks_for_student(3)

同样的原则可以应用于count_ranges函数，只需编写一个通用函数并将标记作为参数传递：

def count_ranges(marks):
    range_counts = [0] * 12
    for num in marks[:]: #change number to select number of draws
        which_range=int(num//5)
        range_counts[which_range] = range_counts1[which_range] + 1
    return range_counts

现在使用它：

def get_range_counts():
    range_counts = []
    student_nums = [1,2,3,4,5] #or range(1,6)
    for n in student_nums:
        marks = open_marks_for_student(n)
        range_counts.append(count_ranges(marks)) #create histogram, append to list
    return range_counts

仍然可以进行一些优化，例如：只打开一次标记文件，而不是每个学生打开一次，或者删除marks中循环中count_ranges的切片（如果您计划修改原始列表，则只需要复制一份副本）循环），但这应该足以让你开始。

在Python中循环以减少重复

1 个答案: