如何获得所有在熊猫中重复的链接元素(行)?

Question

我正在尝试获得所有已被客户消耗的食物id。如果我给一个输入，让我们说一个customerId = C15，那么它应该返回所有的food_id的客户有一个链接。我试过了。

import datetime
import random
import pandas as pd
pd.set_option('display.max_rows', None)
pd.set_option('display.max_columns', None)
df = pd.DataFrame({
'food_id':['B1', 'CF', 'C1', 'CF', 'B1', 'IO', 'B1', 'FT', 'BR','CF', 'C1', 'SM', 'IC','B1', 'SM'],
'purch_amt':[150.5, 270.65, 65.26, 110.5, 948.5, 2400.6, 5760, 1983.43, 2480.4, 250.45, 75.29, 3045.6, 500, 500, 250],
'ord_date': ['05-10-2022','09-10-2022','05-10-2022','08-17-2022','10-09-2022','07-27-2022','10-09-2022','10-10-2022','10-10-2022','06-17-2022','07-08-2022','04-25-2022','02-08-2022','08-08-2022','07-09-2022'],
'customer_id':['C31','C31','D35','D31','C35','D31','C35','D31','D35','C31','D35','D35','D35','C35','C38']})

def get_mostly_ordered_food_id():
print('Max food_id ordered')
result = df['food_id'].value_counts().rename_axis('food_id').reset_index(name='counts')
df2 = pd.DataFrame(result)
print(df2.head(3))
def show_data():
result = df.groupby(['customer_id'])
print(result.first())
def test_add_data():
foodID = ['B1', 'CF', 'C1', 'IO', 'FT', 'BR', 'IC', 'SM']
customerID = ['C31','D35','D31','C35','C38']
date = 'Test-Data'
for i in range(0,10):
df.loc[len(df.index)] = [random.choice(customerID), random.choice(foodID), date, 500]
print(df)
get_mostly_ordered_food_id()

如果我给出一个输入，比如一个顾客ID，那么它应该返回所有被消耗的食物ID。我也不应该重复。

试图获得例如输入"C15"(即客户id)

它应该返回例如"B1, C2"(这是食物id)没有重复。(如果有任何重复，它应该抛出一个警告。)

(注意:关于这份作业的更多内容可以在这个相关问题中阅读。)

Answer 1

def foods_of_customer(df, customer_id):
foods = df['food_id'].loc[df['customer_id'] == customer_id]
return foods.values

方法df.loc[]返回其中满足条件的所有行。food.values方法将数据帧转换为NunPy数组。之后，您可以应用np.unique以避免重复。