如何遍歷字典中每個鍵具有多個值的值

Question

我有一本這樣的字典：

{4722: "['children', 'dance', 'education','technology', 'teaching']",
3200: "['alternative energy', 'sustainability', 'technology']",
1636: "['computers', 'performance', 'simplicity', 'software', 'ice']",
1697: "['MacArthur grant', 'inequality', 'technology', 'pollution']"}

現在，我想在每一行中找到“技術”一詞，並對鍵求和。 像這里一樣，我的總和為4722 + 3200 + 1697。

誰能幫我嗎？

我應該提到我的原始數據框有2000行。

Answer 1

使用內置函數sum() ，並傳遞一個適當的生成器表達式： sum(k for k,v in d.items() if 'technology' in v) （nb在Python2中使用d.iteritems() ）。

可運行的演示：

d = {
    4722: "['children', 'dance', 'education','technology', 'teaching']",
    3200: "['alternative energy', 'sustainability', 'technology']",
    1636: "['computers', 'performance', 'simplicity', 'software', 'ice']",
    1697: "['MacArthur grant', 'inequality', 'technology', 'pollution']"
}

result = sum(k for k,v in d.items() if 'technology' in v)
assert result == 9619

參考文獻：

• https://docs.python.org/2/library/functions.html#sum
• https://docs.python.org/2/library/stdtypes.html#dict.items
• https://docs.python.org/2/reference/expressions.html#grammar-token-generator_expression

Answer 2

your_data = {
    4722: "['children', 'dance', 'education','technology', 'teaching']",
    3200: "['alternative energy', 'sustainability', 'technology']",
    1636: "['computers', 'performance', 'simplicity', 'software', 'ice']",
    1697: "['MacArthur grant', 'inequality', 'technology', 'pollution']"
}

sum_up = 0
for k, v in your_data.items():
    if 'technology' in v:
        sum_up += k

print('sum_up:', sum_up)