Question

如何在numpy数组的第一列中具有相等值的行之间求和？例如：

removeIf

非常感谢任何帮助。

Answer 1

Pandas具有非常强大的groupby功能，这使得这非常简单。

import pandas as pd

n = np.array([[1,2,3],
             [1,4,6], 
             [2,3,5],
             [2,6,2],
             [3,4,8]])

df = pd.DataFrame(n, columns = ["First Col", "Second Col", "Third Col"])

df.groupby("First Col").sum()

Answer 2

方法＃1

这是基于np.bincount -

的numpythonic矢量化方式

# Initial setup             
N = A.shape[1]-1
unqA1, id = np.unique(A[:, 0], return_inverse=True)

# Create subscripts and accumulate with bincount for tagged summations
subs = np.arange(N)*(id.max()+1) + id[:,None]
sums = np.bincount( subs.ravel(), weights=A[:,1:].ravel() )

# Append the unique elements from first column to get final output
out = np.append(unqA1[:,None],sums.reshape(N,-1).T,1)

示例输入，输出 -

In [66]: A
Out[66]: 
array([[1, 2, 3],
       [1, 4, 6],
       [2, 3, 5],
       [2, 6, 2],
       [7, 2, 1],
       [2, 0, 3]])

In [67]: out
Out[67]: 
array([[  1.,   6.,   9.],
       [  2.,   9.,  10.],
       [  7.,   2.,   1.]])

方法＃2

这是另一个基于np.cumsum和np.diff -

的内容

# Sort A based on first column
sA = A[np.argsort(A[:,0]),:]

# Row mask of where each group ends
row_mask = np.append(np.diff(sA[:,0],axis=0)!=0,[True])

# Get cummulative summations and then DIFF to get summations for each group
cumsum_grps = sA.cumsum(0)[row_mask,1:]
sum_grps = np.diff(cumsum_grps,axis=0)

# Concatenate the first unique row with its counts
counts = np.concatenate((cumsum_grps[0,:][None],sum_grps),axis=0)

# Concatenate the first column of the input array for final output
out = np.concatenate((sA[row_mask,0][:,None],counts),axis=1)

基准

这是针对问题 -

目前提出的基于numpy的方法的一些运行时测试

In [319]: A = np.random.randint(0,1000,(100000,10))

In [320]: %timeit cumsum_diff(A)
100 loops, best of 3: 12.1 ms per loop

In [321]: %timeit bincount(A)
10 loops, best of 3: 21.4 ms per loop

In [322]: %timeit add_at(A)
10 loops, best of 3: 60.4 ms per loop

In [323]: A = np.random.randint(0,1000,(100000,20))

In [324]: %timeit cumsum_diff(A)
10 loops, best of 3: 32.1 ms per loop

In [325]: %timeit bincount(A)
10 loops, best of 3: 32.3 ms per loop

In [326]: %timeit add_at(A)
10 loops, best of 3: 113 ms per loop

似乎Approach #2: cumsum + diff表现得非常好。

Answer 3

尝试使用pandas。按第一列分组，然后按行进行求和。像

这样的东西

df.groupby(df.ix[:,1]).sum()

Answer 4

在朋友np.unique和np.add.at的帮助下：

>>> unq, unq_inv = np.unique(A[:, 0], return_inverse=True)
>>> out = np.zeros((len(unq), A.shape[1]), dtype=A.dtype)
>>> out[:, 0] = unq
>>> np.add.at(out[:, 1:], unq_inv, A[:, 1:])

>>> out  # A was the OP's array
array([[1, 6, 9],
       [2, 9, 7],
       [3, 4, 8]])

对列中值相等的行求和

4 个答案:

基准