我正在嘗試創建串列“付款”的總和，該串列在代碼中可以正常作業。如果它看起來像這樣，我如何迭代/獲取相同串列的總和：
disbursements = [(2000, datetime.strptime('01-12-22', '%d-%m-%y')),
(1000, datetime.strptime('01-12-22', '%d-%m-%y')),]
我只想遍歷整數 2000,1000。
def loan(principal, interest_rate):
disbursements = [2000,1000]
total_loan = principal sum(disbursements)
print(f"Total loan amount is {total_loan} EUR.")
payback = int(total_loan (total_loan * (interest_rate / 100)))
print(f"Total amount to pay back is {payback} EUR.")
principal = 10000
interest_rate = 10
loan(principal, interest_rate)
uj5u.com熱心網友回復：
disbursements = [
(2000, datetime.strptime('01-12-22', '%d-%m-%y')),
(1000, datetime.strptime('01-12-22', '%d-%m-%y'))
]
所以你想遍歷 each 的第一個值tuple，對吧？
for amount_, date_ in disbursements:
... # Put here the body of the iteration
在這種情況下，我認為您只需要sum金額，對嗎？
amounts_ = [amount_ for amount_, date in disbursements]
amounts = sum(amounts_)
或者，以更簡潔有效的方式：
amounts = sum(amount_ for amount_, date in disbursements)
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標籤：Python列表循环
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