Question

基于数据框ds的选择d，其中包含：

{ 'x': d.x, 'y': d.y, 'a':d.a, 'b':d.b, 'c':d.c 'row:d.n'})

n行有x行，范围从0到n-1。列n是必需的，因为它是一个选择，需要保留索引以供以后查询。

如何有效地计算每列（a_0, a_1, etc）的每一行（例如a, b, c）之间的差异，而不会丢失行信息（例如，新列中包含已使用行的索引）？

MWE

样本选择ds：

             x           y      a     b      c     n

    554.607085  400.971878   9789  4151   6837   146
    512.231450  405.469524   8796  3811   6596   225
    570.427284  694.369140   1608  2019   2097   291

期望的输出：

dist欧几里德距离math.hypot(x2 - x1, y2 - y1)

da, db, dc

的

da: np.abs(a1-a2)

ns一个字符串，其中包含n个已使用的行

结果如下：

             dist          da        db       dc         ns
42.61365102824963         993       340      241    146-225
293.82347069813255       8181      2132     4740    146-291
                ..         ..        ..       ..    225-291

Answer 1

您可以使用itertools.combinations()生成对：

首先读取数据：

import pandas as pd
from io import StringIO
import numpy as np

text = """             x           y      a     b      c     n
    554.607085  400.971878   9789  4151   6837   146
    512.231450  405.469524   8796  3811   6596   225
    570.427284  694.369140   1608  2019   2097   291"""

df = pd.read_csv(StringIO(text), delim_whitespace=True)

创建索引并计算结果：

from itertools import combinations

index = np.array(list(combinations(range(df.shape[0]), 2)))

df1, df2 = [df.iloc[idx].reset_index(drop=True) for idx in index.T]

res = pd.concat([
    np.hypot(df1.x - df2.x, df1.y - df2.y),
    df1[["a", "b", "c"]] - df2[["a", "b", "c"]],
    df1.n.astype(str) + "-" + df2.n.astype(str)
], axis=1)

res.columns = ["dist", "da", "db", "dc", "ns"]
res

输出：

         dist    da    db    dc       ns
0   42.613651   993   340   241  146-225
1  293.823471  8181  2132  4740  146-291
2  294.702805  7188  1792  4499  225-291

Answer 2

这种方法很好地利用了Pandas和潜在的numpy功能，但矩阵操作有点难以跟踪：

import pandas as pd, numpy as np

ds = pd.DataFrame(
    [
        [554.607085, 400.971878,  9789, 4151,  6837,  146], 
        [512.231450, 405.469524,  8796, 3811,  6596,  225],
        [570.427284, 694.369140,  1608, 2019,  2097,  291]
    ],
    columns = ['x', 'y', 'a', 'b', 'c', 'n']
)

def concat_str(*arrays):
    result = arrays[0]
    for arr in arrays[1:]:
        result = np.core.defchararray.add(result, arr)
    return result

# Make a panel with one item for each column, with a square data frame for 
# each item, showing the differences between all row pairs.
# This creates perpendicular matrices of values based on the underlying numpy arrays;
# then numpy broadcasts them along the missing axis when calculating the differences
p = pd.Panel(
    (ds.values[np.newaxis,:,:] - ds.values[:,np.newaxis,:]).transpose(), 
    items=['d'+c for c in ds.columns], major_axis=ds.index, minor_axis=ds.index
)
# calculate euclidian distance
p['dist'] = np.hypot(p['dx'], p['dy'])
# create strings showing row relationships
p['ns'] = concat_str(ds['n'].values.astype(str)[:,np.newaxis], '-', ds['n'].values.astype(str)[np.newaxis,:])
# remove unneeded items
del p['dx'], p['dy'], p['dn']
# convert to frame
diffs = p.to_frame().reindex_axis(['dist', 'da', 'db', 'dc', 'ns'], axis=1)
diffs

这给出了：

                   dist    da    db    dc       ns
major minor                                       
0     0        0.000000     0     0     0  146-146
      1       42.613651   993   340   241  146-225
      2      293.823471  8181  2132  4740  146-291
1     0       42.613651  -993  -340  -241  225-146
      1        0.000000     0     0     0  225-225
      2      294.702805  7188  1792  4499  225-291
2     0      293.823471 -8181 -2132 -4740  291-146
      1      294.702805 -7188 -1792 -4499  291-225
      2        0.000000     0     0     0  291-291

计算所有可能行的差异

2 个答案: