mik*_*123 6 python lambda pandas
这个问题与我之前发布的问题非常相似，只有一处更改。我不仅想计算所有列的绝对差异，还想找到“Z”列的幅度差异，因此如果当前 Z 比前一个大 1.1 倍，则保留它。
（问题的更多背景）
df = pd.DataFrame({
'rank': [1, 1, 2, 2, 3, 3],
'x': [0, 3, 0, 3, 4, 2],
'y': [0, 4, 0, 4, 5, 5],
'z': [1, 3, 1.2, 3.25, 3, 6],
})
print(df)
# rank x y z
# 0 1 0 0 1.00
# 1 1 3 4 3.00
# 2 2 0 0 1.20
# 3 2 3 4 3.25
# 4 3 4 5 3.00
# 5 3 2 5 6.00
这就是我想要的输出
output = pd.DataFrame({
'rank': [1, 1, 2, 3],
'x': [0, 3, 0, 2],
'y': [0, 4, 0, 5],
'z': [1, 3, 1.2, 6],
})
print(output)
# rank x y z
# 0 1 0 0 1.0
# 1 1 3 4 3.0
# 2 2 0 0 1.2
# 5 3 2 5 6.00
基本上我想要发生的是，如果前一个排名有任何带有 x, y （+- 1 双向）和 z (<1.1z) 的行来删除它。
因此，对于排名 1 的行，排名 2 中的任意行具有 x = (-1-1)、y = (-1-1)、z= (<1.1) 或 x = (2-5)、y 的任意组合= (3-5), z= (<3.3) 我希望将其删除
这是使用numpy 广播的解决方案：
\n# Initially, no row is dropped\ndf[\'drop\'] = False\n\nfor r in range(df[\'rank\'].min(), df[\'rank\'].max()):\n # Find the x_min, x_max, y_min, y_max, z_max of the current rank\n cond = df[\'rank\'] == r\n x, y, z = df.loc[cond, [\'x\',\'y\',\'z\']].to_numpy().T\n x_min, x_max = x + [[-1], [1]] # use numpy broadcasting to \xc2\xb11 in one command\n y_min, y_max = y + [[-1], [1]]\n z_max = z * 1.1\n\n # Find the x, y, z of the next rank. Raise them one dimension\n # so that we can make a comparison matrix again x_min, x_max, ...\n cond = df[\'rank\'] == r + 1\n if not cond.any():\n continue\n x, y, z = df.loc[cond, [\'x\',\'y\',\'z\']].to_numpy().T[:, :, None]\n\n # Condition to drop a row\n drop = (\n (x_min <= x) & (x <= x_max) &\n (y_min <= y) & (y <= y_max) &\n (z <= z_max)\n ).any(axis=1)\n df.loc[cond, \'drop\'] = drop\n\n# Result\ndf[~df[\'drop\']]\n一个更精简的版本（并且可能更快）。这是让你未来的队友在阅读代码时感到困惑的好方法：
\nr, x, y, z = df[[\'rank\', \'x\', \'y\', \'z\']].T.to_numpy()\nrr, xx, yy, zz = [col[:,None] for col in [r, x, y, z]]\n\ndrop = (\n (rr == r + 1) &\n (x-1 <= xx) & (xx <= x+1) &\n (y-1 <= yy) & (yy <= y+1) &\n (zz <= z*1.1)\n).any(axis=1)\n\n# Result\ndf[~drop]\n其作用是将每一行相互比较df（包括其自身），并在以下情况下返回 True（即丢弃）：
rank== 另一行rank + 1；和x, y, z落在其他行的指定范围内x, y, z我修改了mozway的功能，使其能够按照您的要求工作。
# comparing 'equal' float values, may go wrong, that's why I am using this constant
DELTA=0.1**12
def check_previous_group(rank, d, groups):
if not rank-1 in groups.groups:
# check if a previous group exists, else flag all rows False (i.e. not to be dropped)
#return pd.Series(False, index=d.index)
return pd.Series(False, index=d.index)
else:
# get previous group (rank-1)
d_prev = groups.get_group(rank-1)
# get the absolute difference per row with the whole dataset
# of the previous group: abs(d_prev-s)
# if differences in x and y are within 1 and z < 1.1*x
# for at least one row of the previous group
# then flag the row to be dropped (True)
return d.apply(lambda s: (abs(d_prev-s)[['x', 'y']].le([1,1]).all(1)&
(s['z']<1.1*d_prev['x']-DELTA)).any(), axis=1)
测试，
>>> df = pd.DataFrame({
'rank': [1, 1, 2, 2, 3, 3],
'x': [0, 3, 0, 3, 4, 2],
'y': [0, 4, 0, 4, 5, 5],
'z': [1, 3, 1.2, 3.25, 3, 6],
})
>>> df
rank x y z
0 1 0 0 1.00
1 1 3 4 3.00
2 2 0 0 1.20
3 2 3 4 3.25
4 3 4 5 3.00
5 3 2 5 6.00
>>> groups = df.groupby('rank')
>>> mask = pd.concat([check_previous_group(rank, d, groups) for rank,d in groups])
>>> df[~mask]
rank x y z
0 1 0 0 1.0
1 1 3 4 3.0
2 2 0 0 1.2
5 3 2 5 6.0
>>> df = pd.DataFrame({
'rank': [1, 1, 2, 2, 3, 3],
'x': [0, 3, 0, 3, 4, 2],
'y': [0, 4, 0, 4, 5, 5],
'z': [1, 3, 1.2, 3.3, 3, 6],
})
>>> df
rank x y z
0 1 0 0 1.0
1 1 3 4 3.0
2 2 0 0 1.2
3 2 3 4 3.3
4 3 4 5 3.0
5 3 2 5 6.0
>>> groups = df.groupby('rank')
>>> mask = pd.concat([check_previous_group(rank, d, groups) for rank,d in groups])
>>> df[~mask]
rank x y z
0 1 0 0 1.0
1 1 3 4 3.0
2 2 0 0 1.2
3 2 3 4 3.3
5 3 2 5 6.0