根據python中的條件拆分字典

我有兩本字典，我想根據以下條件將我的產品字典拆分為串列：

product = {
    'k1': 30.99999999999994,
    'k2': 400.0,
    'k3': 50.0,
    'k4': 400.00000000000006,
    'k5': 400.0,
    'k6': 50.0,
    'k7': 60.0,
    'k8': 300.0,
    'k9': 40.0}

types = {
    'k1': 2,
    'k2': 1,
    'k3': 1,
    'k4': 1,
    'k5': 1,
    'k6': 1,
    'k7': 1,
    'k8': 1,
    'k9': 2}

• 1 - 如果一個產品的價值小于 200 應該加入其他產品，而新的產品串列的價值小于 600
• 2 - 不同型別的產品無法加入

我已經撰寫了以下代碼，但它在字典的最后一個索引上有一個錯誤，我每次都得到錯誤的結果。

k = list(product.keys())

ProductValue = []
ProductName = []
for idx, name in enumerate(k):
    value = product[name]
    if idx == 0:
        temp1 = [value]
        temp2 = [name]
        continue
    if idx == len(k) - 1:
        ProductValue.append(temp1)
        ProductName.append(temp2)
        continue
    if value < 200 or product[k[idx - 1]] < 200:
        sec1 = types[k[idx - 1]]
        sec2 = types[k[idx]]
        if ((sum(temp1)   value) > 600) or (sec1 != sec2):
            ProductValue.append(temp1)
            ProductName.append(temp2)
            temp1 = [value]
            temp2 = [name]
            continue
        else:
            temp1.append(value)
            temp2.append(name)

    else:
        ProductValue.append(temp1)
        ProductName.append(temp2)
        temp1 = [value]
        temp2 = [name]

我從這段代碼中得到的是：

ProductValue = [[30.99999999999994], [400.0, 50.0], [400.00000000000006], [400.0, 50.0, 60.0], [300.0]]
ProductName = [['k1'], ['k2', 'k3'], ['k4'], ['k5', 'k6', 'k7'], ['k8']]

缺少最后一個值為 40 的產品。我嘗試修復我的代碼，但我讓它變得更糟

uj5u.com熱心網友回復：

編輯： 請按照添加的評論修復錯誤并洗掉條件idx==len(k)-1

product = {
    'k1': 30.99999999999994,
    'k2': 400.0,
    'k3': 50.0,
    'k4': 400.00000000000006,
    'k5': 400.0,
    'k6': 50.0,
    'k7': 60.0,
    'k8': 300.0,
    'k9': 40.0}

types = {
    'k1': 2,
    'k2': 1,
    'k3': 1,
    'k4': 1,
    'k5': 1,
    'k6': 1,
    'k7': 1,
    'k8': 1,
    'k9': 2}


k = list(product.keys())

ProductValue = []
ProductName = []
# create empty sublists
temp1 = []
temp2 = []
for idx, name in enumerate(k):
    value = product[name]
    if idx == 0:
        temp1 = [value]
        temp2 = [name]
        continue
    if value < 200 or product[k[idx - 1]] < 200:
        sec1 = types[k[idx - 1]]
        sec2 = types[k[idx]]
        if ((sum(temp1)   value) > 600) or (sec1 != sec2):
            ProductValue.append(temp1)
            ProductName.append(temp2)
            temp1 = [value]
            temp2 = [name]
            continue
        else:
            temp1.append(value)
            temp2.append(name)

    else:
        # append temp1 and temp1 when for condition fails
        ProductValue.append(temp1)
        ProductName.append(temp2)
        temp1 = [value]
        temp2 = [name]
else:
    # append the last element to the output list
    ProductValue.append(temp1)
    ProductName.append(temp2)


print(ProductName)
print(ProductValue)

輸出：

[(['k1'], [30.99999999999994]),
 (['k2', 'k3'], [400.0, 50.0]),
 (['k4'], [400.00000000000006]),
 (['k5', 'k6', 'k7'], [400.0, 50.0, 60.0]),
 (['k8'], [300.0]),
 (['k9'], [40.0])]

輸入2：

product = {
    'k1': 30.99999999999994,
    'k2': 400.0,
    'k3': 50.0,
    'k4': 400.00000000000006,
    'k5': 400.0,
    'k6': 50.0,
    'k7': 60.0,
    'k8': 300.0,
    'k9': 40.0}

types = {
    'k1': 2,
    'k2': 1,
    'k3': 1,
    'k4': 1,
    'k5': 1,
    'k6': 1,
    'k7': 1,
    'k8': 1,
    'k9': 1}

輸出：

[(['k1'], [30.99999999999994]),
 (['k2', 'k3'], [400.0, 50.0]),
 (['k4'], [400.00000000000006]),
 (['k5', 'k6', 'k7'], [400.0, 50.0, 60.0]),
 (['k8', 'k9'], [300.0, 40.0])]

標籤：Python 列表 字典

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