Python 应用于数据帧的*交错*组
我有一个3轴数据的DataFrames,带有一个成员标签,用于分组:Python 应用于数据帧的*交错*组,python,pandas,numpy,dataframe,scipy,Python,Pandas,Numpy,Dataframe,Scipy,我有一个3轴数据的DataFrames,带有一个成员标签,用于分组: df = pd.DataFrame( [[0, 1, 2, 0], [-1, 0, 1, 0], [-2, 0, 3, 1], [1, 1, 3, 1], [1, 0, 2, 2], [1, 0, 3, 2],
df = pd.DataFrame( [[0, 1, 2, 0],
[-1, 0, 1, 0],
[-2, 0, 3, 1],
[1, 1, 3, 1],
[1, 0, 2, 2],
[1, 0, 3, 2],
[6, 2, 1, 5],
[-4, 3, 0, 5],
[1, 0, -1, 6],
[0, 0, 3, 6]], columns = ['x', 'y', 'z', 'member'])
我的目标有点矫揉造作:我希望找到每个组的点与下一个n\u skip
组之间的成对距离,从最小到最大排序。这个n_skip
就是我所说的交错:
例如,对于n_skip=2
,我希望找到以下距离:
member==0的行与
member==1、2的行相对
member==1的行与
member==2,5的行相对
member==2的行与
member==5、6的行相对
member==5的行与
member==6的行相对
- 不计算
成员==6
apply
来并行化数据帧上的函数。将函数应用于交错组的快速方法是什么
EDIT1 我的解决方案(仅适用于一个轴):
## Heading ### Organize by group membership
groups = df.groupby('member')
# Define constants
max_member = 6
n_skip = 2
start_row = 0
matrix = np.zeros((df.shape[0], df.shape[0]))
# Iterate for each group
for i in range(max_member):
try:
pts_curr = groups.get_group(i)
except KeyError:
continue
# Save end row index
end_row = start_row + pts_curr.shape[0]
# Save start col index
start_col = end_row
# Grab the destination group nodes
for j in range(i+1, int(np.min([i+n_skip+1, max_member]))):
try:
pts_clr_next = groups.get_group(j)
except KeyError:
continue
# Save end col index
end_col = start_col + pts_clr_next.shape[0]
# Calculate cdist
z_sq = cdist(pts_curr[['z']], pts_next[['z']])
# Save results in matrix at right positions
matrix[start_row:end_row, start_col:end_col] = z_sq
# update col index
start_col = end_col
# update row index
start_row = end_row
对4K行进行交叉合并并不太糟糕(产生约1600万行)。让我们尝试交叉合并和查询:
n = 2
# dummy key
df['dummy'] = 1
# this is the member group number
df['rank'] = df['member'].rank(method='dense')
# cross merge and filter
new_df = (df.merge(df, on='dummy')
.query('rank_x<rank_y<=rank_x+@n')
)
# euclidean distance
dist = (new_df[['x_x','y_x','z_x']].sub(new_df[['x_y','y_y','z_y']].values)**2).sum(1)**.5
# output dataframe with member label
pd.DataFrame({'member1':new_df['member_x'], 'member2':new_df['member_y'],
'dist':dist})
选项2:如果有大数据帧,循环可能不会太糟糕:
from scipy.spatial.distance import cdist
ret = []
for i in set(df['rank']):
this_group = df['rank']==i
other_groups = df['rank'].between(i,i+n, inclusive=False)
t = df.loc[this_group,['x','y','z']].values
o = df.loc[other_groups,['x','y','z']].values
ret.append(cdist(t,o).ravel())
dist = np.concatenate(ret)
成员
组是否总是具有相同的行数?不,不幸的是,它们没有。为了方便起见,我制作了一个人为的示例,使每个组拥有相同数量的成员资格。您的数据有多大?也许交叉合并就可以了。它不是很大。实际上,我正在构建一个图,上面的矩阵
用于计算邻接矩阵。我的图有大约4000个节点;但是我有很多这样的独立图(~1000个图)。然而,我正在努力缩短任何ms
的时间。以前我对交叉合并的经验是,它相当慢。df.query('rank\ux
from scipy.spatial.distance import cdist
ret = []
for i in set(df['rank']):
this_group = df['rank']==i
other_groups = df['rank'].between(i,i+n, inclusive=False)
t = df.loc[this_group,['x','y','z']].values
o = df.loc[other_groups,['x','y','z']].values
ret.append(cdist(t,o).ravel())
dist = np.concatenate(ret)