我有大量的小熊猫数据帧，我必须在这些数据帧上进行完整的行比较，并将结果写入新的数据帧，这些数据帧将在以后连接。
对于行比较，我正在使用iloc在数据帧的长度上进行双循环。我不知道是否有更快的方式，我这样做的方式似乎很慢：
# -*- coding: utf-8 -*-
import pandas as pd
import time
def processFrames1(DF):
LL = []
for i in range(len(DF)):
for j in range(len(DF)):
if DF.iloc[i][0] != DF.iloc[j][0]:
T = {u'T1':DF.iloc[i][0]}
T[u'T2'] = DF.iloc[j][0]
T[u'T3'] = 1
if DF.iloc[i][2] > DF.iloc[j][2]:
T[u'T4'] = 1
elif DF.iloc[i][2] < DF.iloc[j][2]:
T[u'T4'] = -1
else:
T[u'T4'] = 0
if DF.iloc[i][1] < DF.iloc[j][1]:
T[u'T5'] = 1
else:
T[u'T5'] = -1
LL.append(T)
return pd.DataFrame.from_dict(LL)
D = [{'A':'XA','B':1,'C':1.4}\
,{'A':'RT','B':2,'C':10}\
,{'A':'HO','B':3,'C':34}\
,{'A':'NJ','B':4,'C':0.41}\
,{'A':'WF','B':5,'C':114}\
,{'A':'DV','B':6,'C':74}\
,{'A':'KP','B':7,'C':2.4}]
P = pd.DataFrame.from_dict(D)
time0 = time.time()
for i in range(10):
X = processFrames1(P)
print time.time()-time0
print X
产生结果：
0.836999893188
T1 T2 T3 T4 T5
0 XA RT 1 -1 1
1 XA HO 1 -1 1
2 XA NJ 1 1 1
3 XA WF 1 -1 1
4 XA DV 1 -1 1
5 XA KP 1 -1 1
6 RT XA 1 1 -1
7 RT HO 1 -1 1
8 RT NJ 1 1 1
9 RT WF 1 -1 1
10 RT DV 1 -1 1
11 RT KP 1 1 1
12 HO XA 1 1 -1
13 HO RT 1 1 -1
14 HO NJ 1 1 1
15 HO WF 1 -1 1
16 HO DV 1 -1 1
17 HO KP 1 1 1
18 NJ XA 1 -1 -1
19 NJ RT 1 -1 -1
20 NJ HO 1 -1 -1
21 NJ WF 1 -1 1
22 NJ DV 1 -1 1
23 NJ KP 1 -1 1
24 WF XA 1 1 -1
25 WF RT 1 1 -1
26 WF HO 1 1 -1
27 WF NJ 1 1 -1
28 WF DV 1 1 1
29 WF KP 1 1 1
30 DV XA 1 1 -1
31 DV RT 1 1 -1
32 DV HO 1 1 -1
33 DV NJ 1 1 -1
34 DV WF 1 -1 -1
35 DV KP 1 1 1
36 KP XA 1 1 -1
37 KP RT 1 -1 -1
38 KP HO 1 -1 -1
39 KP NJ 1 1 -1
40 KP WF 1 -1 -1
41 KP DV 1 -1 -1
使用这个代表性的数据帧只需要10次，我需要花费超过一百万的时间。
有没有更快的方法来进行那些完整的行比较？
EDIT1： 经过一些修改后，我可以让Javier的代码创建正确的输出：
def compare_values1(x,y):
if x>y: return 1
elif x<y: return -1
else: return 0
def compare_values2(x,y):
if x<y: return 1
elif x>y: return -1
else: return 0
def processFrames(P):
D = P.to_dict(orient='records')
d_A2B = {d["A"]:d["B"] for d in D}
d_A2C = {d["A"]:d["C"] for d in D}
keys = list(d_A2B.keys())
LL = []
for i in range(len(keys)):
k_i = keys[i]
for j in range(len(keys)):
if i != j:
k_j = keys[j]
LL.append([k_i,k_j,1,compare_values1(\
d_A2C[k_i],d_A2C[k_j]),compare_values2(d_A2B[k_i],d_A2B[k_j])])
return pd.DataFrame(LL,columns=['T1','T2','T3','T4','T5'])
此功能的工作速度提高了约60倍。
EDIT2： 四种可能性的最终判决：
===============使用小数据帧：
我原来的功能：
%timeit processFrames1(P)
10 loops, best of 3: 85.3 ms per loop
%timeit processFrames2(P)
1 loop, best of 3: 286 ms per loop
Javier的修改代码：
%timeit processFrames3(P)
1000 loops, best of 3: 1.24 ms per loop
Divakar的方法：
%timeit processFrames4(P)
1000 loops, best of 3: 1.98 ms per loop
===============对于大型数据帧：
我原来的功能：
%timeit processFrames1(P)
1 loop, best of 3: 2.22 s per loop
%timeit processFrames2(P)
1 loop, best of 3: 295 ms per loop
Javier的修改代码：
%timeit processFrames3(P)
100 loops, best of 3: 3.13 ms per loop
Divakar的方法：
%timeit processFrames4(P)
100 loops, best of 3: 2.19 ms per loop
所以这是最后两个之间的关系。感谢大家的帮助，非常需要加速。
编辑3：
Divakar编辑了他们的代码，这是新结果：
小数据框：
%timeit processFrames(P)
1000 loops, best of 3: 492 µs per loop
大型数据框：
%timeit processFrames(P)
1000 loops, best of 3: 844 µs per loop
非常令人印象深刻，绝对是赢家。
编辑4：
Divakar的方法略有修改，因为我现在正在我的程序中使用它：
def processFrames(P):
N = len(P)
N_range = np.arange(N)
valid_mask = (N_range[:,None] != N_range).ravel()
colB = P.B.values
colC = P.C.values
T2_arr = np.ones(N*N,dtype=int)
T4_arr = np.zeros((N,N),dtype=int)
T4_arr[colC[:,None] > colC] = 1
T4_arr[colC[:,None] < colC] = -1
T5_arr = np.zeros((N,N),dtype=int)
T5_arr[colB[:,None] > colB] = -1
T5_arr[colB[:,None] < colB] = 1
strings = P.A.values
c0,c1 = np.meshgrid(strings,strings)
arr = np.column_stack((c1.ravel(), c0.ravel(), T2_arr,T4_arr.ravel(),\
T5_arr.ravel()))[valid_mask]
return arr[:,0],arr[:,1],arr[:,2],arr[:,3],arr[:,4]
我正在创建一个包含五个键的字典，每个列表示五个结果列，然后我只是用结果扩展列表，一旦我完成，我就从字典中制作一个pandas数据帧。这比连接现有数据帧要快得多。
PS：我从中学到了一件事：如果你能以任何方式避免使用iloc，就不要使用iloc。答案 0 :(得分：3)
以下是使用NumPy broadcasting -
def processFrames1_broadcasting(P):
N = len(P)
N_range = np.arange(N)
valid_mask = (N_range[:,None] != N_range).ravel()
colB = P.B.values
colC = P.C.values
T2_arr = np.ones(N*N,dtype=int)
T4_arr = np.zeros((N,N),dtype=int)
T4_arr[colC[:,None] > colC] = 1
T4_arr[colC[:,None] < colC] = -1
T5_arr = np.where(colB[:,None] < colB,1,-1)
strings = P.A.values
c0,c1 = np.meshgrid(strings,strings)
arr = np.column_stack((c1.ravel(), c0.ravel(), T2_arr,T4_arr.ravel(),\
T5_arr.ravel()))[valid_mask]
df = pd.DataFrame(arr, columns=[['T1','T2','T3','T4','T5']])
return df
运行时测试 -
对于问题中发布的样本，我最终得到的运行时间是 -
In [337]: %timeit processFrames1(P)
10 loops, best of 3: 93.1 ms per loop
In [338]: %timeit processFrames1_jezrael(P) #@jezrael's soln
10 loops, best of 3: 74.8 ms per loop
In [339]: %timeit processFrames1_broadcasting(P)
1000 loops, best of 3: 561 µs per loop
答案 1 :(得分：1)
不要使用熊猫。使用词典并保存：
def compare_values(x,y):
if x>y: return 1
elif x<y: return -1
else: return 0
def processFrames(P):
d_A2B = dict(zip(P["A"],P["B"]))
d_A2C = dict(zip(P["A"],P["C"]))
keys = list(d_A2B.keys())
d_ind2key = dict(zip(range(len(keys)),keys))
LL = []
for i in range(len(keys)):
k_i = keys[i]
for j in range(i+1,len(keys)):
k_j = keys[j]
c1 = compare_values(d_A2C[k_i],d_A2C[k_j])
c2 = -compare_values(d_A2B[k_i],d_A2B[k_j])
LL.append([k_i,k_j,1,c1,c2])
LL.append([k_j,k_i,1,-c1,-c2])
return pd.DataFrame(LL,columns=['T1','T2','T3','T4','T5'])
答案 2 :(得分：1)
您可以使用：
#cross join
P['one'] = 1
df = pd.merge(P,P, on='one')
df = df.rename(columns={'A_x':'T1','A_y':'T2'})
#remove duplicates
df = df[df.T1 != df.T2]
df.reset_index(drop=True, inplace=True)
#creates new columns
df['T3'] = 1
df['T4'] = (df.C_x > df.C_y).astype(int).replace({0:-1})
df['T5'] = (df.B_x < df.B_y).astype(int).replace({0:-1})
#remove other columns by subset
df = df[['T1','T2','T3','T4','T5']]
print (df)
T1 T2 T3 T4 T5
0 XA RT 1 -1 1
1 XA HO 1 -1 1
2 XA NJ 1 1 1
3 XA WF 1 -1 1
4 XA DV 1 -1 1
5 XA KP 1 -1 1
6 RT XA 1 1 -1
7 RT HO 1 -1 1
8 RT NJ 1 1 1
9 RT WF 1 -1 1
10 RT DV 1 -1 1
11 RT KP 1 1 1
12 HO XA 1 1 -1
13 HO RT 1 1 -1
14 HO NJ 1 1 1
15 HO WF 1 -1 1
16 HO DV 1 -1 1
17 HO KP 1 1 1
18 NJ XA 1 -1 -1
19 NJ RT 1 -1 -1
20 NJ HO 1 -1 -1
21 NJ WF 1 -1 1
22 NJ DV 1 -1 1
23 NJ KP 1 -1 1
24 WF XA 1 1 -1
25 WF RT 1 1 -1
26 WF HO 1 1 -1
27 WF NJ 1 1 -1
28 WF DV 1 1 1
29 WF KP 1 1 1
30 DV XA 1 1 -1
31 DV RT 1 1 -1
32 DV HO 1 1 -1
33 DV NJ 1 1 -1
34 DV WF 1 -1 -1
35 DV KP 1 1 1
36 KP XA 1 1 -1
37 KP RT 1 -1 -1
38 KP HO 1 -1 -1
39 KP NJ 1 1 -1
40 KP WF 1 -1 -1
41 KP DV 1 -1 -1
<强>的时间设置：
In [339]: %timeit processFrames1(P)
10 loops, best of 3: 44.2 ms per loop
In [340]: %timeit jez(P1)
10 loops, best of 3: 43.3 ms per loop
如果使用你的时间：
time0 = time.time()
for i in range(10):
X = processFrames1(P)
print (time.time()-time0)
0.4760475158691406
time0 = time.time()
for i in range(10):
X = jez(P1)
print (time.time()-time0)
0.4400441646575928
测试代码：
P1 = P.copy()
def jez(P):
P['one'] = 1
df = pd.merge(P,P, on='one')
df = df.rename(columns={'A_x':'T1','A_y':'T2'})
df = df[df.T1 != df.T2]
df.reset_index(drop=True, inplace=True)
df['T3'] = 1
df['T4'] = (df.C_x > df.C_y).astype(int).replace({0:-1})
df['T5'] = (df.B_x < df.B_y).astype(int).replace({0:-1})
df = df[['T1','T2','T3','T4','T5']]
return (df)
def processFrames1(DF):
LL = []
for i in range(len(DF)):
for j in range(len(DF)):
if DF.iloc[i][0] != DF.iloc[j][0]:
T = {u'T1':DF.iloc[i][0]}
T[u'T2'] = DF.iloc[j][0]
T[u'T3'] = 1
if DF.iloc[i][2] > DF.iloc[j][2]:
T[u'T4'] = 1
elif DF.iloc[i][2] < DF.iloc[j][2]:
T[u'T4'] = -1
else:
T[u'T4'] = 0
if DF.iloc[i][1] < DF.iloc[j][1]:
T[u'T5'] = 1
else:
T[u'T5'] = -1
LL.append(T)
return pd.DataFrame.from_dict(LL)
EDIT1：
我尝试用5倍更大的dataFrame测试：
D = [{'A':'XA','B':1,'C':1.4}\
,{'A':'RB','B':2,'C':10}\
,{'A':'HC','B':3,'C':34}\
,{'A':'ND','B':4,'C':0.41}\
,{'A':'WE','B':5,'C':114}\
,{'A':'DF','B':6,'C':74}\
,{'A':'KG','B':7,'C':2.4}\
,{'A':'XH','B':1,'C':1.4}\
,{'A':'RI','B':2,'C':10}\
,{'A':'HJ','B':3,'C':34}\
,{'A':'NK','B':4,'C':0.41}\
,{'A':'WL','B':5,'C':114}\
,{'A':'DM','B':6,'C':74}\
,{'A':'KN','B':7,'C':2.4}\
,{'A':'XO','B':1,'C':1.4}\
,{'A':'RP','B':2,'C':10}\
,{'A':'HQ','B':3,'C':34}\
,{'A':'NR','B':4,'C':0.41}\
,{'A':'WS','B':5,'C':114}\
,{'A':'DT','B':6,'C':74}\
,{'A':'KU','B':7,'C':2.4}\
,{'A':'XV','B':1,'C':1.4}\
,{'A':'RW','B':2,'C':10}\
,{'A':'HX','B':3,'C':34}\
,{'A':'NY','B':4,'C':0.41}\
,{'A':'WZ','B':5,'C':114}\
,{'A':'D1','B':6,'C':74}\
,{'A':'K2','B':7,'C':2.4}\
,{'A':'X3','B':1,'C':1.4}\
,{'A':'R4','B':2,'C':10}\
,{'A':'H5','B':3,'C':34}\
,{'A':'N6','B':4,'C':0.41}\
,{'A':'W7','B':5,'C':114}\
,{'A':'D8','B':6,'C':74}\
,{'A':'K9','B':7,'C':2.4} ]
P = pd.DataFrame.from_dict(D)
P1 = P.copy()
time0 = time.time()
for i in range(10):
X = processFrames1(P)
print (time.time()-time0)
12.230222940444946
time0 = time.time()
for i in range(10):
X = jez(P1)
print (time.time()-time0)
0.4440445899963379
In [351]: %timeit processFrames1(P)
1 loop, best of 3: 1.21 s per loop
In [352]: %timeit jez(P1)
10 loops, best of 3: 43.7 ms per loop