我有以下类型的数据帧-
数据框
A B C
5 10 15
20 25 30
我想做以下手术-
A_B A_C B_C
-0.33 -0.5 -0.2
-0.11 -0.2 -0.09
A_B,A_C,B_C对应于-
A_B: A-B/A+B
A_C: A-C/A+C
B_C: B-C/B+C
我用的是-
colnames = df.columns.tolist()[:-1]
list_name=[]
for i,c in enumerate(colnames):
if i!=len(colnames):
for k in range(i+1,len(colnames)):
df[c+'_'+colnames[k]]=(df[c]-
df[colnames[k]])/(df[c]+df[colnames[k]])
list_name.append(c+'_'+colnames[k])
但问题是我的实际数据帧的大小是
5*381
形状,因此A_B, A_C and so on
的实际组合数是5*72390
形状,这需要60分钟才能运行。所以我试着把它转换成numpy数组,这样我就可以用Numba优化它来有效地计算它(Parallel programming approach to solve pandas problems),但是我不能把它转换成numpy数组。
此外,任何其他解决这一问题的办法也受到欢迎。
最佳答案
使用:
df = pd.DataFrame({
'A':[5,20],
'B':[10,25],
'C':[15,30]
})
print (df)
A B C
0 5 10 15
1 20 25 30
首先将列的所有组合获取到两个列表(
a
表示元组的第一个值,b
表示第二个值):from itertools import combinations
a, b = zip(*(combinations(df.columns, 2)))
然后按列表对重复列使用
DataFrame.loc
:df1 = df.loc[:, a]
print (df1)
A A B
0 5 5 10
1 20 20 25
df2 = df.loc[:, b]
print (df2)
B C C
0 10 15 15
1 25 30 30
将值转换为最终数据帧的numpy数组,并通过列表理解获取新列名:
c = [f'{x}_{y}' for x, y in zip(a, b)]
arr1 = df1.values
arr2 = df2.values
df = pd.DataFrame((arr1-arr2)/(arr1+arr2), columns=c)
print (df)
A_B A_C B_C
0 -0.333333 -0.5 -0.200000
1 -0.111111 -0.2 -0.090909
另一种解决方案非常相似,只需按列长度创建组合,最后通过索引创建新列名称:
from itertools import combinations
a, b = zip(*(combinations(np.arange(len(df.columns)), 2)))
arr = df.values
cols = df.columns.values
arr1 = arr[:, a]
arr2 = arr[:, b]
c = [f'{x}_{y}' for x, y in zip(cols[np.array(a)], cols[np.array(b)])]
df = pd.DataFrame((arr1-arr2)/(arr1+arr2), columns=c)
性能:
在5行381列中测试:
np.random.seed(2019)
df = pd.DataFrame(np.random.randint(10,100,(5,381)))
df.columns = ['c'+str(i+1) for i in range(df.shape[1])]
#print (df)
In [4]: %%timeit
...: a, b = zip(*(combinations(np.arange(len(df.columns)), 2)))
...: arr = df.values
...: cols = df.columns.values
...: arr1 = arr[:, a]
...: arr2 = arr[:, b]
...: c = [f'{x}_{y}' for x, y in zip(cols[np.array(a)], cols[np.array(b)])]
...: pd.DataFrame((arr1-arr2)/(arr1+arr2), columns=c)
...:
62 ms ± 7.29 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [5]: %%timeit
...: a, b = zip(*(combinations(df.columns, 2)))
...: df1 = df.loc[:, a]
...: df2 = df.loc[:, b]
...: arr1 = df1.values
...: arr2 = df2.values
...: c = [f'{x}_{y}' for x, y in zip(a, b)]
...: pd.DataFrame((arr1-arr2)/(arr1+arr2), columns=c)
...:
63.2 ms ± 253 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [7]: %%timeit
...: func1(df)
...:
89.2 ms ± 331 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [8]: %%timeit
...: a, b = zip(*(combinations(df.columns, 2)))
...: df1 = df.loc[:, a]
...: df2 = df.loc[:, b]
...: c = [f'{x}_{y}' for x, y in zip(a, b)]
...: pd.DataFrame((df1.values-df2.values)/(df1.values+df2.values), columns=c)
...:
69.8 ms ± 6.04 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
关于python - 通过列组合提高算术运算的性能,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/55116552/