当只知道前任和后继 id 时确定元素 id 的巧妙方法 - Python

Question

我遇到了以下情况：
我的输入数据包含几个具有前任和后继 id 的元素 - 但不是它自己的 id。
如下表所示

+----------+-------------+-------------+
| Element  | NextBlockID | PrevBlockID |
+----------+-------------+-------------+
| Block623 | c308002017  | 2a08003017  |
+----------+-------------+-------------+
| Block625 | 0           | c308002017  |
+----------+-------------+-------------+
| Block622 | 2808003017  | c208002017  |
+----------+-------------+-------------+
| Block620 | c208002017  | 0           |
+----------+-------------+-------------+
| Block621 | 2a08003017  | be08003017  |
+----------+-------------+-------------+
| Block624 | 2908002017  | 2808003017  |
+----------+-------------+-------------+

现在我想确定每个元素的元素 id 并将其添加为新列。
我现在正在做的是确定 PrevBlockID 为零的第一个元素。然后查找我的第一个元素的 NextBlockID 与另一个元素的 PrevBlockID 匹配的位置，并将其添加到具有 Block622、Block624、... 的所有渴望元素 ID 的列表中，直到不再匹配 NextBlockID。
然后我查看 NextBlockID = 0 的元素。这是最后一个元素。如果它的值 PrevBlockID 与 NextBlockID 匹配，我会一个一个地获取 uneager 元素 ID（Block623、Block621）

所以之后我想要一个这样的输出表

+----------+-------------+-------------+------------+
| Element  | NextBlockID | PrevBlockID | ElementID  |
+----------+-------------+-------------+------------+
| Block623 | c308002017  | 2a08003017  | 2808003017 |
+----------+-------------+-------------+------------+
| Block625 | 0           | c308002017  | 2908002017 |
+----------+-------------+-------------+------------+
| Block622 | 2808003017  | c208002017  | 2a08003017 |
+----------+-------------+-------------+------------+
| Block620 | c208002017  | 0           | be08003017 |
+----------+-------------+-------------+------------+
| Block621 | 2a08003017  | be08003017  | c208002017 |
+----------+-------------+-------------+------------+
| Block624 | 2908002017  | 2808003017  | c308002017 |
+----------+-------------+-------------+------------+

我的输入数据存储在熊猫数据框中。是否有任何更智能/更快的解决方案，然后逐个迭代这些值？

2020 年 6 月 14 日下午 2:07 更新： 抱歉，这里的混淆是我到目前为止得到的代码：

import pandas as pd
f1 = 'DF_DetermineBlockID.csv'
df = pd.read_csv(f1, sep=';')
Ids = pd.Series([], dtype=object)
df = df.sort_values("PrevBlock")  # sorted to get 0 value in first pos
df.index = pd.RangeIndex(len(df.index))

successor = df[df.index == 0].squeeze()["NextBlock"]
Ids = Ids.append(pd.Series(successor, index=[1]))
a = 1
PrevBlockFound = not df[df["PrevBlock"] == successor].empty

while PrevBlockFound:
    a += 2
    successor = df[df["PrevBlock"] == successor].squeeze()["NextBlock"]
    Ids = Ids.append(pd.Series(successor, index=[a]))
    PrevBlockFound = not df[df["PrevBlock"] == successor].empty

predecessor = df[df["NextBlock"] == "0"].squeeze()["PrevBlock"]
a -= 1
Ids = Ids.append(pd.Series(predecessor, index=[a]))
NextBlockFound = not df[df["NextBlock"] == predecessor].empty

while NextBlockFound:
    a -= 2
    predecessor = df[df["NextBlock"] == predecessor].squeeze()["PrevBlock"]
    Ids = Ids.append(pd.Series(predecessor, index=[a]))
    NextBlockFound = not df[df["NextBlock"] == predecessor].empty

df = pd.merge(df, Ids.rename('BlockID'), left_index=True, right_index=True)

Answer 1

找出您正在寻找的逻辑有点困难 - 但这段代码会产生您期望的输出：

df = df.sort_values("Element")
df["ElementID"] = df.PrevBlockID.shift(-1)
df.loc[df.ElementID.isna(), "ElementID"] = df.NextBlockID.shift()
df.sort_index(inplace = True)

输出是：

    Element NextBlockID PrevBlockID   ElementID
0  Block623  c308002017  2a08003017  2808003017
1  Block625           0  c308002017  2908002017
2  Block622  2808003017  c208002017  2a08003017
3  Block620  c208002017           0  be08003017
4  Block621  2a08003017  be08003017  c208002017
5  Block624  2908002017  2808003017  c308002017