我正在尝试找出在熊猫数据帧上执行搜索和排序的最快方法。以下是我试图完成的工作之前和之后的数据帧。
以前：
flightTo flightFrom toNum fromNum toCode fromCode
ABC DEF 123 456 8000 8000
DEF XYZ 456 893 9999 9999
AAA BBB 473 917 5555 5555
BBB CCC 917 341 5555 5555
搜索/排序后：
flightTo flightFrom toNum fromNum toCode fromCode
ABC XYZ 123 893 8000 9999
AAA CCC 473 341 5555 5555
在这个例子中，我基本上是在试图过滤掉存在于最终目的地之间的"航班"。这应该通过使用某种删除重复方法来完成，但让我感到困惑的是如何处理所有列。二叉搜索是实现这一目标的最佳方式吗？暗示赞赏，努力弄清楚这一点。
可能的边缘情况：
如果数据被切换并且我们的终端连接在同一列中怎么办？
flight1 flight2 1Num 2Num 1Code 2Code
ABC DEF 123 456 8000 8000
XYZ DEF 893 456 9999 9999
搜索/排序后：
flight1 flight2 1Num 2Num 1Code 2Code
ABC XYZ 123 893 8000 9999
从逻辑上讲，这种情况不应该发生。毕竟，你怎么能去DEF-ABC和DEF-XYZ？你不能，但"端点"仍然是ABC-XYZ
这是
网络问题，所以我们用networkx，注意，这里可以有两个以上的停止，这意味着你可以有一些类似NY-DC-WA-NC的情况
import networkx as nx
G=nx.from_pandas_edgelist(df, 'flightTo', 'flightFrom')
# create the nx object from pandas dataframe
l=list(nx.connected_components(G))
# then we get the list of components which as tied to each other ,
# in a net work graph , they are linked
L=[dict.fromkeys(y,x) for x, y in enumerate(l)]
# then from the above we can create our map dict ,
# since every components connected to each other ,
# then we just need to pick of of them as key , then map with others
d={k: v for d in L for k, v in d.items()}
# create the dict for groupby , since we need _from as first item and _to as last item
grouppd=dict(zip(df.columns.tolist(),['first','last']*3))
df.groupby(df.flightTo.map(d)).agg(grouppd) # then using agg with dict yield your output
Out[22]:
flightTo flightFrom toNum fromNum toCode fromCode
flightTo
0 ABC XYZ 123 893 8000 9999
1 AAA CCC 473 341 5555 5555
安装networkx
- 点：
pip install networkx - 蟒蛇：
conda install -c anaconda networkx
这是一个 NumPy 解决方案，在性能相关的情况下可能很方便：
def remove_middle_dest(df):
x = df.to_numpy()
# obtain a flat numpy array from both columns
b = x[:,0:2].ravel()
_, ix, inv = np.unique(b, return_index=True, return_inverse=True)
# Index of duplicate values in b
ixs_drop = np.setdiff1d(np.arange(len(b)), ix)
# Indices to be used to replace the content in the columns
replace_at = (inv[:,None] == inv[ixs_drop]).argmax(0)
# Col index of where duplicate value is, 0 or 1
col = (ixs_drop % 2) ^ 1
# 2d array to index and replace values in the df
# index to obtain values with which to replace
keep_cols = np.broadcast_to([3,5],(len(col),2))
ixs = np.concatenate([col[:,None], keep_cols], 1)
# translate indices to row indices
rows_drop, rows_replace = (ixs_drop // 2), (replace_at // 2)
c = np.empty((len(col), 5), dtype=x.dtype)
c[:,::2] = x[rows_drop[:,None], ixs]
c[:,1::2] = x[rows_replace[:,None], [2,4]]
# update dataframe and drop rows
df.iloc[rows_replace, 1:] = c
return df.drop(rows_drop)
建议的数据帧产生预期的输出：
print(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 DEF XYZ 456 893 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555
remove_middle_dest(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 893 8000 9999
2 AAA CCC 473 341 5555 5555
这种方法不假定重复项所在的行有任何特定的顺序，这同样适用于列(以涵盖问题中描述的边缘情况(。例如，如果我们使用以下数据帧：
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC DEF 123 456 8000 8000
1 XYZ DEF 893 456 9999 9999
2 AAA BBB 473 917 5555 5555
3 BBB CCC 917 341 5555 5555
remove_middle_dest(df)
flightTo flightFrom toNum fromNum toCode fromCode
0 ABC XYZ 123 456 8000 9999
2 AAA CCC 473 341 5555 5555