Question

我正在尝试获取数据帧八列之间每种可能组合的计数（所有行值为1）。基本上我需要了解有多少次不同的重叠存在。

我尝试使用itertools.product来获取所有组合，但似乎不起作用。

import pandas as pd
import numpy as np
import itertools

df = pd.read_excel('filename.xlsx')

df.head(15)

    a   b   c   d   e   f   g   h
0   1   0   0   0   0   1   0   0
1   1   0   0   0   0   0   0   0
2   1   0   1   1   1   1   1   1
3   1   0   1   1   0   1   1   1
4   1   0   0   0   0   0   0   0
5   0   1   0   0   1   1   1   1
6   1   1   0   0   1   1   1   1
7   1   1   1   1   1   1   1   1
8   1   1   0   0   1   1   0   0
9   1   1   1   0   1   0   1   0
10  1   1   1   0   1   1   0   0
11  1   0   0   0   0   1   0   0
12  1   1   1   1   1   1   1   1
13  1   1   1   1   1   1   1   1
14  0   1   1   1   1   1   1   0


print(list(itertools.product(new_df.columns)))

预期输出将是一个数据帧，其中包含每个有效组合的行数（n）（其中行中的值均为1）。

例如：

会给

combination   count

a              12
a_b             7
b               9

请注意，输出将需要包含a和h之间的所有可能组合，而不仅仅是成对

Answer 1

Powerset组合

使用powerset食谱，

s = pd.Series({
    '_'.join(c): df[c].min(axis=1).sum() 
    for c in map(list, filter(None, powerset(df)))
})

a                  13
b                   9
c                   8
d                   6
e                  10
f                  12
g                   9
h                   7
a_b                 7
...

成对组合

这是特例，可以向量化。

from itertools import combinations

u = df.T.dot(df)   
pd.DataFrame({
    'combination': [*map('_'.join, combinations(df, 2))], 
    # pandas < 0.24
    # 'count': u.values[np.triu_indices_from(u, k=1)]
    # pandas >= 0.24
    'count': u.to_numpy()[np.triu_indices_from(u, k=1)]
})

您可以使用dot，然后提取上三角矩阵值：

  combination  count
0         a_b      7
1         a_c      7
2         a_d      5
3         a_e      8
4         a_f     10
5         a_g      7
6         a_h      6
7         b_c      6
8         b_d      4
9         b_e      9

Answer 2

如果只有值1和0，则可以执行以下操作：

df= pd.DataFrame({
'a': [1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1],
'b': [1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0],
'c': [1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1],
'd': [1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1],
})

(df.a * df.b).sum()

结果为4。

要获取所有组合，可以使用combinations中的itertools：

from itertools import combinations

analyze=[(col,) for col in df.columns]
analyze.extend(combinations(df.columns, 2))
for cols in analyze:
    num_ser= None
    for col in cols:
        if num_ser is None:
            num_ser= df[col]
        else:
            num_ser*= df[col]
    num= num_ser.sum()
    print(f'{cols} contains {num}')

结果是：

('a',) contains 4
('b',) contains 7
('c',) contains 11
('d',) contains 23
('a', 'b') contains 4
('a', 'c') contains 4
('a', 'd') contains 4
('b', 'c') contains 7
('b', 'd') contains 7
('c', 'd') contains 11

Answer 3

共同矩阵是您所需要的：

让我们首先构造一个示例：

import numpy as np
import pandas as pd

mat = np.zeros((5,5))
mat[0,0] = 1
mat[0,1] = 1
mat[1,0] = 1
mat[2,1] = 1
mat[3,3] = 1
mat[3,4] = 1
mat[2,4] = 1
cols = ['a','b','c','d','e']
df = pd.DataFrame(mat,columns=cols)
print(df)

     a    b    c    d    e
0  1.0  1.0  0.0  0.0  0.0
1  1.0  0.0  0.0  0.0  0.0
2  0.0  1.0  0.0  0.0  1.0
3  0.0  0.0  0.0  1.0  1.0
4  0.0  0.0  0.0  0.0  0.0

现在我们构造共现矩阵：

# construct the cooccurence matrix:
co_df = df.T.dot(df)
print(co_df)

     a    b    c    d    e
a  2.0  1.0  0.0  0.0  0.0
b  1.0  2.0  0.0  0.0  1.0
c  0.0  0.0  0.0  0.0  0.0
d  0.0  0.0  0.0  1.0  1.0
e  0.0  1.0  0.0  1.0  2.0

最后需要的结果：

result = {}

for c1 in cols:
    for c2 in cols:
        if c1 == c2:
            if c1 not in result:
                result[c1] = co_df[c1][c2]
        else:

            if '_'.join([c1,c2]) not in result:
                result['_'.join([c1,c2])] = co_df[c1][c2]


print(result)



{'a': 2.0, 'a_b': 1.0, 'a_c': 0.0, 'a_d': 0.0, 'a_e': 0.0, 'b_a': 1.0, 'b': 2.0, 'b_c': 0.0, 'b_d': 0.0, 'b_e': 1.0, 'c_a': 0.0, 'c_b': 0.0, 'c': 0.0, 'c_d': 0.0, 'c_e': 0.0, 'd_a': 0.0, 'd_b': 0.0, 'd_c': 0.0, 'd': 1.0, 'd_e': 1.0, 'e_a': 0.0, 'e_b': 1.0, 'e_c': 0.0, 'e_d': 1.0, 'e': 2.0}

Answer 4

当您碰巧有8列时，np.packbits和 np.bincount在这里很方便：

import numpy as np
import pandas as pd

# make large example
ncol, nrow = 8, 1_000_000
df = pd.DataFrame(np.random.randint(0,2,(nrow,ncol)), columns=list("abcdefgh"))

from time import time
T = [time()]
# encode as binary numbers and count
counts = np.bincount(np.packbits(df.values.astype(np.uint8)),None,256)

# find sets in other sets
rng = np.arange(256, dtype=np.uint8)
contained = (rng & rng[:, None]) == rng[:, None]

# and sum
ccounts = (counts * contained).sum(1)

# if there are empty bins, remove them
nz = np.where(ccounts)[0].astype(np.uint8)

# helper to build bin labels 
a2h = np.array(list("abcdefgh"))

# put labels to counts
result = pd.Series(ccounts[nz], index = ["_".join((*a2h[np.unpackbits(i).view(bool)],)) for i in nz])



from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

T.append(time())
s = pd.Series({
    '_'.join(c): df[c].min(axis=1).sum() 
    for c in map(list, filter(None, powerset(df)))
})
T.append(time())

print("packbits {:.3f} powerset {:.3f}".format(*np.diff(T)))
print("results equal", (result.sort_index()[1:]==s.sort_index()).all())

这与powerset方法产生的结果相同，但实际上要快1000倍：

packbits 0.016 powerset 21.974
results equal True

数据框列之间所有可能组合的计数

4 个答案:

Powerset组合

成对组合