通过递增/递减最后找到的值来向前/向后填充na?

The result should be something like this in the end (here is a copy of it):+----+---------+--------+------------------------+----------------------+-----------------+| | Values | Signal | forward signal rows nr | backward signal rows | value at signal |+----+---------+--------+------------------------+----------------------+-----------------+| 0 | 1420.49 | | | | || 1 | 1421.12 | | | | || 2 | 1418.95 | | | | || 3 | 1419.04 | 1 | 1 | 4 | 1416.97 || 4 | 1419.04 | | 2 | 3 | 1416.97 || 5 | 1417.51 | | 3 | 2 | 1416.97 || 6 | 1416.97 | | 4 | 1 | 1416.97 || 7 | 1413.21 | -1 | -1 | -3 | 1412.57 || 8 | 1411.49 | | -2 | -2 | 1412.57 || 9 | 1412.57 | | -3 | -1 | 1412.57 || 10 | 1408.55 | 1 | 1 | 3 | 1413.38 || 11 | 1409.16 | | 2 | 2 | 1413.38 || 12 | 1413.38 | | 3 | 1 | 1413.38 || 13 | 1413.38 | 1 | 1 | 5 | 1397.62 || 14 | 1402.35 | | 2 | 4 | 1397.62 || 15 | 1397.8 | | 3 | 3 | 1397.62 || 16 | 1398.36 | | 4 | 2 | 1397.62 || 17 | 1397.62 | | 5 | 1 | 1397.62 || 18 | 1394.58 | -1 | -1 | -3 | 1399.9 || 19 | 1399.05 | | -2 | -2 | 1399.9 || 20 | 1399.9 | | -3 | -1 | 1399.9 || 21 | 1398.96 | -1 | -1 | -5 | 1398.66 || 22 | 1398.96 | | -2 | -4 | 1398.66 || 23 | 1393.69 | | -3 | -3 | 1398.66 || 24 | 1398.13 | | -4 | -2 | 1398.66 || 25 | 1398.66 | | -5 | -1 | 1398.66 || 26 | 1398.02 | 1 | 1 | 4 | 1398.13 || 27 | 1397.97 | | 2 | 3 | 1398.13 || 28 | 1396.05 | | 3 | 2 | 1398.13 || 29 | 1398.13 | | 4 | 1 | 1398.13 |+----+---------+--------+------------------------+----------------------+-----------------+我通过几个嵌套循环获得了最终结果，但是问题是它们在几百万行的较大数据帧上效率很低. I achieved the final result with a few nested loops but the problem is that they are very inefficient on larger data frames of a few million rows. 推荐答案基于信号的分组的常用方法(IMHO应该对此提供更好的本机支持)使用compare-cumsum-groupby模式.这里的比较是确定信号条目是否为空，之后我们进行累加和，以便每个信号组都有自己的ID(组ID或GID).剩下的只是算术.The usual approach to signal-based groupings (which we should really have better native support for, IMHO) to use the compare-cumsum-groupby pattern. Here the comparison is to determine whether a signal entry is null or not, after which we do a cumulative sum so that each signal group has its own id (group id, or gid). The rest is just arithmetic.虽然这里有些重复，我们可以重构，但我感到很懒，所以:While there's some duplication here we could refactor away, I'm feeling lazy, and so:gid = df["Signal"].notnull().cumsum()dg = df.groupby(gid)sign = dg["Signal"].transform("first")df["forward signal rows"] = (dg.cumcount() + 1) * signdf["backward signal rows"] = (dg["Signal"].transform("size") - dg.cumcount()) * signdf["value at signal"] = dg["Values"].transform("last")df.loc[gid == 0, "value at signal"] = np.nan为我提供了一个与您的目标相匹配的框架.gives me a frame matching your target one. 这篇关于通过递增/递减最后找到的值来向前/向后填充na?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！