英文:
Lists manipulation with indexs
问题
我有两个列表。
列表 A:
A = ["apple", "cherry", "pear", "mango", "banana", "grape", "kiwi", "orange", "pineapple"]
列表 B:
B = [{"offset": 0, "xx": 789}, {"offset": 3, "xx": 921}, {"offset": 6, "xx": 89}]
思路是使用 B 中每个项目的偏移量作为结果数组中的 "xx" 值的索引偏移量。
例如，这将是预期结果：
C = [{"fruit": "apple", "xx": 789},{"fruit": "cherry", "xx": 789},{"fruit": "pear", "xx": 789},{"fruit": "mango", "xx": 921},{"fruit": "banana", "xx": 921},{"fruit": "grape", "xx": 921},{"fruit": "kiwi", "xx": 89},{"fruit": "orange", "xx": 89},{"fruit": "pineapple", "xx": 89}]
例如，B[0] 的 "offset" 为 0，这意味着 C 的索引 >= 0 将具有 "xx" 值等于 B[0]['xx']。然后，B[0]['offset'] 为 3，将为索引 >= 3 的 C 项目设置新的 "xx" 值，依此类推。
我可以使用数据框和 pandas 实现类似的结果。但由于 pandas 库相当庞大，我被要求在不使用 pandas 的情况下完成。
英文:
I have two lists.
List A :
A = ["apple","cherry","pear","mango","banana","grape","kiwi","orange","pineapple"]
List B :
B = [{"offset":0, "xx":789},{"offset":3, "xx":921},{"offset":6, "xx":89}]
The idea is to use the offset from each item in B as an index offset for setting the xx values in our results array.
For instance, this would be the expected result:
C=[{"fruit":"apple","xx":789},{"fruit":"cherry","xx":789},{"fruit":"pear","xx":789},{"fruit":"mango","xx":921},{"fruit":"banana","xx":921},{"fruit":"grape","xx":921},{"fruit":"kiwi","xx":89},{"fruit":"orange","xx":89},{"fruit":"pineapple","xx":89},]
For example, B[0] has "offset" of 0. this means that C of index >= 0 will have an "xx" value of B[0]['xx']. Then we have B[0]['offset'] of 3 that will set new "xx" values to the C items with index >= 3 and so on.
I am able to acheive a similar result using a dataframes and pandas. But since pandas library is quite heavy, I am requested to do it without using pandas.
答案1
得分: 6
使用一个简单的循环如何？
# 将 B 重新格式化dic = {d['offset']: d['xx'] for d in B}# {0: 789, 3: 921, 6: 89}C = []v = Nonefor i, a in enumerate(A):v = dic.get(i, v) # 如果达到阈值，更新值C.append({'fruit': a, 'xx': v})print(C)
输出：
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
英文:
What about using a simple loop?
# rework B in a better formatdic = {d['offset']:d['xx'] for d in B}# {0: 789, 3: 921, 6: 89}C = []v = Nonefor i, a in enumerate(A):v = dic.get(i, v) # if we reached a threshold, update the valueC.append({'fruit':a, 'xx': v})print(C)
Output:
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
答案2
得分: 1
如果需要使B的结构如此，你可以这样做：
A = ["apple", "cherry", "pear", "mango", "banana", "grape", "kiwi", "orange", "pineapple"]B = [{"offset": 0, "xx": 789}, {"offset": 3, "xx": 921}, {"offset": 6, "xx": 89}]C = []B_iter = 0for i, fruit in enumerate(A):# 检查是否不是最后一个元素并且下一个元素是新范围的开始if B[B_iter] != B[-1] and B[B_iter + 1]["offset"] == i:B_iter += 1C.append({"fruit": fruit, "xx": B[B_iter]["xx"]})print(C)
输出：
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
英文:
If the structure of B is required to be this way, you can do this:
A = ["apple","cherry","pear","mango","banana","grape","kiwi","orange","pineapple"]B = [{"offset":0, "xx":789},{"offset":3, "xx":921},{"offset":6, "xx":89}]C = []B_iter = 0for i, fruit in enumerate(A):# check if not the last element and next element is start of new rangeif B[B_iter] != B[-1] and B[B_iter+1]["offset"] == i:B_iter += 1C.append({"fruit": fruit, "xx": B[B_iter]["xx"]})print(C)
Output:
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
答案3
得分: 0
如果偏移量始终是3的倍数，您可以简单地进行整数除法来将实际索引映射到偏移量。
A = ["apple","cherry","pear","mango","banana","grape","kiwi","orange","pineapple"]B = [{"offset": 0, "xx": 789}, {"offset": 3, "xx": 921}, {"offset": 6, "xx": 89}]C = [{"fruit": fruit, "xx": B[int(idx/3)]["xx"]} for idx, fruit in enumerate(A)]
输出:
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
英文:
If the offset is always multiple of 3 you can simply do integer division to map the actual index to offset.
A = ["apple","cherry","pear","mango","banana","grape","kiwi","orange","pineapple"]B = [{"offset": 0, "xx": 789}, {"offset": 3, "xx": 921}, {"offset": 6, "xx": 89}]C = [{"fruit": fruit,"xx": B[int(idx/3)]["xx"]} for idx, fruit in enumerate(A)]
Output:
[{'fruit': 'apple', 'xx': 789},{'fruit': 'cherry', 'xx': 789},{'fruit': 'pear', 'xx': 789},{'fruit': 'mango', 'xx': 921},{'fruit': 'banana', 'xx': 921},{'fruit': 'grape', 'xx': 921},{'fruit': 'kiwi', 'xx': 89},{'fruit': 'orange', 'xx': 89},{'fruit': 'pineapple', 'xx': 89}]
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