Python,比较和计算列表中的数据

时间:2017-07-10 08:02:00

标签: python list

我正在尝试计算3门课程的出勤率。

Excel电子表格中的原始数据如下所示(“1”表示有人参加,“0”表示没有):

enter image description here

需要的是计算:

参加A课程的学生中,有多少人(%)上过B课,并参加过课程C. 在参加过B课程的学员中,有多少人(%)参加了A课程,并参加了课程C. 在参加过C课程的学生中,有多少人(%)参加了A课程,并参加了课程B.

我从代码中得到的结果就在这里。

他们的意思是:

在参加过A级课程的学生中,100%参加了A课程,50%参加了B课程,75%参加了课程C.

参加课程B的学生中,67%参加了A课程,100%参加了课程B,100%参加了课程C.

在C课程中,75%参加了A课程,75%参加了B课程,100%参加了课程C.

from xlrd import open_workbook,cellname
import xlwt, xlrd
from xlutils.copy import copy 
from xlwt import Workbook,easyxf,Formula

workbook = xlrd.open_workbook("C:\\Sheet1.xls")
old_sheet = workbook.sheet_by_index(0)

B1 = old_sheet.cell(0, 1).value
C1 = old_sheet.cell(0, 2).value
D1 = old_sheet.cell(0, 3).value

sum_of_Column_B = []
sum_of_Column_C = []
sum_of_Column_D = []

Column_B_B = []
Column_B_C = []
Column_B_D = []

Column_C_B = []
Column_C_C = []
Column_C_D = []

Column_D_B = []
Column_D_C = []
Column_D_D = []


for row_index in range(1, old_sheet.nrows):
#     Column_A = old_sheet.cell(row_index, 0).value
    Column_B = old_sheet.cell(row_index, 1).value
    Column_C = old_sheet.cell(row_index, 2).value
    Column_D = old_sheet.cell(row_index, 3).value

    sum_of_Column_B.append(int(Column_B))
    sum_of_Column_C.append(int(Column_C))
    sum_of_Column_D.append(int(Column_D))

# Paragraph 1
    if Column_B == 1 and Column_B == 1:
        Column_B_B.append(1)    
    if Column_B == 1 and Column_C == 1:
        Column_B_C.append(1)
    if Column_B == 1 and Column_D == 1:
        Column_B_D.append(1)

# Paragraph 2
    if Column_C == 1 and Column_B == 1:
        Column_C_B.append(1)
    if Column_C == 1 and Column_C == 1:
        Column_C_C.append(1)
    if Column_C == 1 and Column_D == 1:
        Column_C_D.append(1)

# Paragraph 3
    if Column_D == 1 and Column_B == 1:
        Column_D_B.append(1)
    if Column_D == 1 and Column_C == 1:
        Column_D_C.append(1)
    if Column_D == 1 and Column_D == 1:
        Column_D_D.append(1)

# Paragraph 1
B_over_B = float(sum(Column_B_B)) / float(sum(sum_of_Column_B))
C_over_B = float(sum(Column_B_C)) / float(sum(sum_of_Column_B))
D_over_B = float(sum(Column_B_D)) / float(sum(sum_of_Column_B))

# Paragraph 2
B_over_C = float(sum(Column_C_B)) / float(sum(sum_of_Column_C))
C_over_C = float(sum(Column_C_C)) / float(sum(sum_of_Column_C))
D_over_C = float(sum(Column_C_D)) / float(sum(sum_of_Column_C))

# Paragraph 3
B_over_D = float(sum(Column_D_B)) / float(sum(sum_of_Column_D))
C_over_D = float(sum(Column_D_C)) / float(sum(sum_of_Column_D))
D_over_D = float(sum(Column_D_D)) / float(sum(sum_of_Column_D))

# Paragraph 1
print B1 + " to " + B1 + " + {0:.0f}%".format(B_over_B * 100)
print C1 + " to " + B1 + " + {0:.0f}%".format(C_over_B * 100)
print D1 + " to " + B1 + " + {0:.0f}%".format(D_over_B * 100)

# Paragraph 2
print " - " * 20
print B1 + " to " + C1 + " + {0:.0f}%".format(B_over_C * 100)
print C1 + " to " + C1 + " + {0:.0f}%".format(C_over_C * 100)
print D1 + " to " + C1 + " + {0:.0f}%".format(D_over_C * 100)

# Paragraph 3
print " - " * 20

print B1 + " to " + D1 + " + {0:.0f}%".format(B_over_D * 100)
print C1 + " to " + D1 + " + {0:.0f}%".format(C_over_D * 100)
print D1 + " to " + D1 + " + {0:.0f}%".format(D_over_D * 100)

正如您所看到的,运行的笨拙代码不是很聪明。如果课程(专栏)的数量大幅增加,例如100列,手动插件是一项繁琐的工作。

进行此类计算的智能方法是什么?谢谢。

{{1}}

2 个答案:

答案 0 :(得分:2)

虚拟数据

import pandas as pd
import itertools
import numpy as np

names = [f'student {i}' for i in range(1, 8)]
courses = [f'course {i}' for i in 'ABC']
df = pd.DataFrame(data = np.random.randint(0, 2, size=(len(names),len(courses))), index = names, columns=courses)

实际上

df = pd.read_excel(filename)
courses = df.columns

您可能需要调整一些参数,尤其是index_colheader

  

DF

    course A    course B    course C
student 1   0   0   1
student 2   1   1   0
student 3   1   0   0
student 4   0   1   0
student 5   1   0   1
student 6   1   0   1
student 7   1   0   1

比较

results = pd.DataFrame(columns=courses, index=courses)
for i, j in itertools.product(courses, repeat=2):
    attended = df[df[i] == 1]
    results.loc[i, j] = sum(attended[i] & attended[j]) / len(attended)
  

结果

    course A    course B    course C
course A    1   0.2     0.6
course B    0.5     1   0
course C    0.75    0   1

因此,参加课程C的人中有75%参加了课程A

答案 1 :(得分:2)

我的csv看起来像这样:

Name;Course A;Course B;Course C
David;1;0;1
Kate;0;1;1  
Tom;1;1;1
Andrew;1;0;0
Jason;0;0;0
Peter;1;1;1

导入这样的数据:

data = pd.read_csv('test.csv',sep=';')
columns = data.columns.tolist ()
columns.remove('Name')

这是一个将课程作为输入并为您提供所需输出的函数:

def assistance(cour):
    print("100 percent of student who assisted {}".format(cour))
    for Course in columns:
        if Course != cour:
            assistance = data.groupby(cour).mean().loc[1, Course] * 100
            print ("assisted {0} at {1} percent".format(Course, assistance))

输出

> assistance('Course A')
100 percent of student who assisted Course A
assisted Course B at 50.0 percent
assisted Course C at 75.0 percent

要在DataFrame中包含所有信息:

df = pd.DataFrame(index=columns, columns=columns)
for row in columns:
    for c in columns:
        if row != c:
            df.loc[row,c] = data.groupby(row).mean().loc[1,c] * 100
        else:
            df.loc[row,c] = float(100)

输出

print(df)

         Course A Course B Course C
Course A      100       50       75
Course B  66.6667      100      100
Course C       75       75      100