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python - 如何在 Django 模型中插入许多（+200）行数据框时提高效率？

问题描述

for transaction in df.itertuples():
    transaction_obj = Transactions()
    transaction_obj.portfolio = strategy.portfolio
    transaction_obj.strategy = strategy
    transaction_obj.transaction_price = transaction.TransactionPrices
    transaction_obj.time_stamp = transaction[0]
    transaction_obj.transaction_type = name
    transaction_obj.brokerage = transaction.Brokerage

    if name == Transactions.LONG:
        if transaction.EntryLong:
            transaction_obj.action = Transactions.BUY
        if transaction.ExitLong:
            transaction_obj.action = Transactions.SELL
    elif name == Transactions.SHORT:
        if transaction.EntryShort:
            transaction_obj.action = Transactions.SELL
        if transaction.ExitShort:
            transaction_obj.action = Transactions.BUY

    transaction_obj.save()

该代码工作正常，没有问题或错误，但是我想减少执行时间。目前，插入 180 行行大约需要 7 秒。我正在使用 PostgreSQL 数据库。

标签： pythondjangopandasdataframedjango-models

解决方案

如果您的脚本假设始终创建记录（而不是修改），那么您可以使用Bulk create优化创建时间：

transactions = []
for transaction in df.itertuples():
    transaction_obj = Transactions()
    transaction_obj.portfolio = strategy.portfolio
    transaction_obj.strategy = strategy
    transaction_obj.transaction_price = transaction.TransactionPrices
    transaction_obj.time_stamp = transaction[0]
    transaction_obj.transaction_type = name
    transaction_obj.brokerage = transaction.Brokerage

    if name == Transactions.LONG:
        if transaction.EntryLong:
            transaction_obj.action = Transactions.BUY
        if transaction.ExitLong:
            transaction_obj.action = Transactions.SELL
    elif name == Transactions.SHORT:
        if transaction.EntryShort:
            transaction_obj.action = Transactions.SELL
        if transaction.ExitShort:
            transaction_obj.action = Transactions.BUY

    transactions.append(transaction_obj)
Transaction.objects.bulk_create(transactions)

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