在Pandas数据框中仅减去匹配索引的值。

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英文:

Subtracting values of only matching indexes in Pandas Dataframes

问题

我一直在编写一个使用Pandas管理商店库存的程序。目前我正在做的是从原始库存中减去订购数量,然后打印交易后的库存。我创建了两个数据框来保存原始库存和用户的购物车。

库存

商品 价格 数量
苹果 1.30 90
香蕉 0.57 100
橙子 1.33 50
15.0 50
记事本 1.0 100

购物车

商品 成本 数量
苹果 13.0 10
橙子 6.65 5

我尝试使用命令 inventory['Quantity'] = inventory['Quantity'] - cart['Quantity'] 来获得最终结果,但结果如下:

商品 价格 数量
苹果 1.30 80
香蕉 0.57 NaN
橙子 1.33 45
15.0 NaN
记事本 1.0 NaN

有人知道如何防止其他行显示为NaN吗?

英文:

I've been working on a program to manage a store inventory using Pandas. For what I am working on now is subtracting the quantity ordered from the original and printing the inventory after the transaction. I created two dataframes to hold the original and the user's cart.

Inventory

Items Price Quantity
Apple 1.30 90
Banana 0.57 100
Orange 1.33 50
Book 15.0 50
Notepad 1.0 100

Cart

Items Cost Quantity
Apples 13.0 10
Orange 6.65 5

I've tried using the command inventory['Quantity'] = inventory['Quantity'] - cart['Quantity'] to get a final result but it comes up as

Items Price Quantity
Apple 1.30 80
Banana 0.57 NaN
Orange 1.33 45
Book 15.0 NaN
Notepad 1.0 NaN

Does anyone know how to stop the other rows from coming back as NaN?

答案1

得分: 0

以下是翻译好的部分:

import pandas as pd

# 创建库存数据框
inventory_data = {'Items': ['苹果', '香蕉', '橙子', '书', '记事本'],
                  '价格': [1.30, 0.57, 1.33, 15.0, 1.0],
                  '数量': [90, 100, 50, 50, 100]}
inventory = pd.DataFrame(inventory_data)

# 创建购物车数据框
cart_data = {'Items': ['苹果', '橙子'],
             '成本': [13.0, 6.65],
             '数量': [10, 5]}
cart = pd.DataFrame(cart_data)

# 将物品名称转换为小写以进行匹配
inventory['Items'] = inventory['Items'].str.lower()
cart['Items'] = cart['Items'].str.lower()

# 在“Items”列上合并库存和购物车数据框
merged = pd.merge(inventory, cart, on='Items', how='left')

# 从库存数量中减去购物车数量
merged['数量'] = merged['数量_x'] - merged['数量_y']

# 用库存数量填充NaN值
merged['数量'] = merged['数量'].fillna(merged['数量_x'])

# 选择必要的列并重命名它们
updated_inventory = merged[['Items', '价格', '数量']].rename(columns={'数量': 'Quantity'})

# 将物品名称转换回其原始大小写
updated_inventory['Items'] = updated_inventory['Items'].str.capitalize()

# 将数量列转换为整数
updated_inventory['数量'] = updated_inventory['数量'].astype(int)

# 打印更新后的库存
print(updated_inventory)

希望这对你有所帮助。

英文:

Use the following approach, and you can solve your problem:

import pandas as pd

# Create the inventory data frame
inventory_data = {'Items': ['Apple', 'Banana', 'Orange', 'Book', 'Notepad'],
                  'Price': [1.30, 0.57, 1.33, 15.0, 1.0],
                  'Quantity': [90, 100, 50, 50, 100]}
inventory = pd.DataFrame(inventory_data)

# Create the cart data frame
cart_data = {'Items': ['Apples', 'Orange'],
             'Cost': [13.0, 6.65],
             'Quantity': [10, 5]}
cart = pd.DataFrame(cart_data)

# Convert item names to lowercase for matching
inventory['Items'] = inventory['Items'].str.lower()
cart['Items'] = cart['Items'].str.lower()

# Merge the inventory and cart data frames on the "Items" column
merged = pd.merge(inventory, cart, on='Items', how='left')

# Subtract cart quantities from inventory quantities
merged['Quantity'] = merged['Quantity_x'] - merged['Quantity_y']

# Fill NaN values with inventory quantities
merged['Quantity'] = merged['Quantity'].fillna(merged['Quantity_x'])

# Select the necessary columns and rename them
updated_inventory = merged[['Items', 'Price', 'Quantity']].rename(columns={'Quantity': 'Quantity'})

# Convert the item names back to their original case
updated_inventory['Items'] = updated_inventory['Items'].str.capitalize()

# Convert the quantity column to integers
updated_inventory['Quantity'] = updated_inventory['Quantity'].astype(int)

# Print the updated inventory
print(updated_inventory)

答案2

得分: 0

使用 Series.subtract,并设置 fill_value = 0,以便非匹配的索引被减去零。

inventory["Quantity"].subtract(cart["Quantity"], fill_value=0)
英文:

Use Series.subtract with fill_value = 0 so that non-matched indices are subtracted by zero.

inventory["Quantity"].subtract(cart["Quantity"], fill_value=0)

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  • 本文由 发表于 2023年6月6日 04:35:07
  • 转载请务必保留本文链接:https://go.coder-hub.com/76409825.html
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