如何对 pandas 数据框进行二进制搜索以查找列值的组合？

Question

对不起，如果这是熊猫文档解释的一个简单问题，但我已经尝试搜索如何做到这一点并且没有任何运气。

我有一个包含几列的 pandas 数据名，我希望能够使用二进制搜索来搜索特定行，因为我的数据集很大，而且我将进行大量搜索。

我的数据如下所示：

Name           Course   Week  Grade
-------------  -------  ----  -----
Homer Simpson  MATH001  1     97
Homer Simpson  MATH001  3     85
Homer Simpson  CSCI100  1     89
John McGuirk   MATH001  2     78
John McGuirk   CSCI100  1     100
John McGuirk   CSCI100  2     96

我希望能够快速搜索我的数据以查找名称、课程和周数的特定组合。名称、课程和周的每个不同组合在数据集中将有 0 行或 1 行。如果我正在搜索的名称、课程和周的组合缺少值，我希望我的搜索返回 0。

例如，我想搜索值(John McGuirk, CSCI100, 1)

是否有内置方法可以做到这一点，还是我必须编写自己的二进制搜索？

更新：

我尝试使用下面一位评论者建议的内置方式执行此操作，并且我还尝试使用为我的特定数据编写的自定义二进制搜索以及使用递归的另一个自定义二进制搜索来执行此操作处理与我的具体示例不同的列。

这些测试的数据框包含 10,000 行。我把时间安排在下面。两种二分搜索的性能都比使用[...] 来获取行要好。我不是 Python 专家，所以我不确定我的代码优化得有多好。

# Load data
from pandas import DataFrame, read_csv
import math
import pandas as pd
import time

file = 'grades.xlsx'
df = pd.read_excel(file)

# This was suggested by one of the commenters below
def get_grade(name, course, week):
    mask = (df.name.values == name) & (df.course.values == course) & (df.week.values == week)
    row = df[mask]
    if row.empty == False:
        return row.grade.values[0]
    else:
        return 0

# Binary search that is specific to my particular data
def get_grade_binary_search(name, course, week):
    lower = 0
    upper = len(df.index) - 1

    while lower <= upper:
        mid = math.floor((lower + upper) / 2)

        row_name = df.iat[mid, 0]            
        if name < row_name:
            upper = mid - 1
        elif name > row_name:
            lower = mid + 1
        else:
            row_course = df.iat[mid, 1]
            if course < row_course:
                upper = mid - 1
            elif course > row_course:
                lower = mid + 1
            else:
                row_week = df.iat[mid, 2]
                if week < row_week:
                    upper = mid - 1
                elif week > row_week:
                    lower = mid + 1
                else:
                    return df.iat[mid, 3]

    return 0    

# General purpose binary search
def get_grade_binary_search_recursive(search_value):
    lower = 0
    upper = len(df.index) - 1

    while lower <= upper:
        mid = math.floor((lower + upper) / 2)

        comparison = compare(search_value, 0, mid)

        if comparison < 0:
            upper = mid - 1
        elif comparison > 0:
            lower = mid + 1
        else:
            return df.iat[mid, len(search_value)]

# Utility method
def compare(search_value, search_column_index, df_value_index):      
    if search_column_index >= len(search_value):
        return 0

    if search_value[search_column_index] < df.iat[df_value_index, search_column_index]:
        return -1
    elif search_value[search_column_index] > df.iat[df_value_index, search_column_index]:
        return 1
    else:
        return compare(search_value, search_column_index + 1, df_value_index)

这是时间安排。我还打印了每次搜索返回值的总和，以验证是否返回了相同的行。

# Non binary search
sum_of_grades = 0
start = time.time()   
for week in range(first_week, last_week + 1):
    for name in names:
        for course in courses:
            val = get_grade(name, course, week)
            sum_of_grades += val                
end = time.time()    
print('elapsed time: ', end - start)
print('sum of grades: ', sum_of_grades)

elapsed time: 26.130020141601562

sum of grades: 498724

# Binary search specific to this data
sum_of_grades = 0
start = time.time()    
for week in range(first_week, last_week + 1):
    for name in names:
        for course in courses:
            val = get_grade_binary_search(name, course, week)
            sum_of_grades += val

end = time.time()    
print('elapsed time: ', end - start)
print('sum of grades: ', sum_of_grades)

elapsed time: 4.4506165981292725

sum of grades: 498724

# Binary search with recursion
sum_of_grades = 0
start = time.time()
for week in range(first_week, last_week + 1):
    for name in names:
        for course in courses:
            val = get_grade_binary_search_recursive([name, course, week])
            sum_of_grades += val           
end = time.time()    
print('elapsed time: ', end - start)
print('sum_of_grades: ', sum_of_grades)

elapsed time: 7.559535264968872

sum_of_grades: 498724

Answer 1

熊猫有searchsorted。

来自备注：

二分查找用于查找需要的插入点。