I have a set of SMPS files that are read by Coin-SMI's SMPS reader, but not by SCIP 6.0's. Actually, SCIP 6.0 shows ''Segmentation fault (core dumped)'' when it tries to read .sto file.
Could you take a quick look at it please to see which part is wrong?
Thank you in advance.
SSLP_3_3_3.cor
NAME SSLP_3_3_3
ROWS
N OBJ
L CON1
L CON2
L CON3
L CON4
E CON5
E CON6
E CON7
COLUMNS
MARKER 'MARKER' 'INTORG'
x[1] OBJ 69.000000 CON1 1.000000
x[1] CON2 -66.000000
x[2] OBJ 42.000000 CON1 1.000000
x[2] CON3 -66.000000
x[3] OBJ 44.000000 CON1 1.000000
x[3] CON4 -66.000000
y[1,1] OBJ -21.000000 CON2 21.000000
y[1,1] CON5 1.000000
y[1,2] OBJ -18.000000 CON3 18.000000
y[1,2] CON5 1.000000
y[1,3] OBJ -11.000000 CON4 11.000000
y[1,3] CON5 1.000000
y[2,1] OBJ -21.000000 CON2 21.000000
y[2,1] CON6 1.000000
y[2,2] OBJ -19.000000 CON3 19.000000
y[2,2] CON6 1.000000
y[2,3] OBJ -4.000000 CON4 4.000000
y[2,3] CON6 1.000000
y[3,1] OBJ -6.000000 CON2 6.000000
y[3,1] CON7 1.000000
y[3,2] OBJ -24.000000 CON3 24.000000
y[3,2] CON7 1.000000
y[3,3] OBJ -8.000000 CON4 8.000000
y[3,3] CON7 1.000000
MARKER 'MARKER' 'INTEND'
y0[1] OBJ 1000.000000 CON2 -1.000000
y0[2] OBJ 1000.000000 CON3 -1.000000
y0[3] OBJ 1000.000000 CON4 -1.000000
RHS
rhs CON1 3.000000 CON2 -0.000000
rhs CON3 -0.000000 CON4 -0.000000
rhs CON5 1.000000 CON6 1.000000
rhs CON7 1.000000
BOUNDS
BV BOUND x[1]
BV BOUND x[2]
BV BOUND x[3]
BV BOUND y[1,1]
BV BOUND y[1,2]
BV BOUND y[1,3]
BV BOUND y[2,1]
BV BOUND y[2,2]
BV BOUND y[2,3]
BV BOUND y[3,1]
BV BOUND y[3,2]
BV BOUND y[3,3]
ENDATA
SSLP_3_3_3.sto
STOCH SSLP_3_3_3
SCENARIOS DISCRETE
SC SCEN1 'ROOT' 0.333333 PERIOD2
rhs CON5 -0.000000
rhs CON6 -0.000000
rhs CON7 -0.000000
SC SCEN2 'ROOT' 0.333333 PERIOD2
rhs CON6 -0.000000
SC SCEN3 'ROOT' 0.333333 PERIOD2
rhs CON7 -0.000000
ENDATA
SSLP_3_3_3.tim
TIME SSLP_3_3_3
PERIODS IMPLICIT
x[1] CON1 PERIOD1
y[1,1] CON2 PERIOD2
ENDATA
SCIP expects rhs to be written upper case, I don't know what the format defines exactly, but this is actually in line with how sections in mps work. We might change this in the future, but until then, just write RHS and everything should work.
Related
I am wondering if there is a way to parallelize leftjoin operations in Julia
For example, I have the following DataFrames :
df1:
Id LDC LDR
a ldc1 ldr1
b ldc2 ldr2
c ldc3 ldr4
d ldc2 ldr3
df2 :
LDC dc1 dc2 dc3
ldc1 0.5 0.4 0.2
ldc2 0.1 0.6 0.7
ldc3 0.4 0.9 0.3
df3
LDR lap1 lap2 lap3
ldr1 0.05 0.06 0.07
ldr2 0.10 0.12 0.13
ldr3 0.01 0.01 0.02
ldr4 0.05 0.06 0.07
I currently make a serial join operation as below
df1 = leftjoin(df1, df2, on = "LDC")
df1 = leftjoin(df1, df3, on = "LDR")
which give me the desired result :
Id LDC LDR dc1 dc2 dc3 lap1 lap2 lap3
a ldc1 ldr1 0.5 0.4 0.2 0.05 0.06 0.07
b ldc2 ldr2 0.1 0.6 0.7 0.10 0.12 0.13
d ldc2 ldr3 0.1 0.6 0.7 0.01 0.01 0.02
c ldc3 ldr4 0.4 0.9 0.3 0.05 0.06 0.07
My question is : Is there a way to "populate" the initial DF (df1) with a parallelized join operation to obtain the same result ?
Thanks for any help you could provide
UPDATE : Here is the code to generate df1,df2 & df3.
df1 = DataFrame(Id = ["a","b","c","d"],
LDC = ["ldc1","ldc2","ldc3","ldc2"],
LDR = ["ldr1","ldr2","ldr4","ldr3"]
)
df2 = DataFrame(LDC = ["ldc1","ldc2","ldc3"],
dc1 = [0.5,0.4,0.2],
dc2 = [0.1,0.6,0.7],
dc3 = [0.4,0.9,0.3]
)
df3 = DataFrame(LDR = ["ldr1","ldr2","ldr3"],
lap1 = [0.05,0.06,0.07],
lap2 = [0.10,0.12,0.13],
lap3 = [0.01,0.01,0.02],
lap4 = [0.05,0.06,0.07]
)
With this dataframe:
item XP_home XP_away
A 0.000000 5.229861
B 6.412500 0.000000
C 5.037361 0.000000
D 0.000000 3.394792
I can sort like so:
df = df.sort_values(by='XP_home', ascending=False).head(2)
and get:
B 6.412500 0.000000
C 5.037361 0.000000
or:
df = df.sort_values(by='XP_away', ascending=False).head(2)
and get:
A 0.000000 5.229861
D 0.000000 3.394792
But how can I sort by the highest of both column values, to get:
item XP_home XP_away
B 6.412500 0.000000
A 0.000000 5.229861
C 5.037361 0.000000
D 0.000000 3.394792
Let us try argsort
out = df.iloc[(-df.filter(like = 'XP').max(1)).argsort()]
item XP_home XP_away
1 B 6.412500 0.000000
0 A 0.000000 5.229861
2 C 5.037361 0.000000
3 D 0.000000 3.394792
You can sort on the max value across rows:
print (df.assign(val=df[["XP_home", "XP_away"]].max(1))
.sort_values("val", ascending=False).drop("val", 1))
item XP_home XP_away
1 B 6.412500 0.000000
0 A 0.000000 5.229861
2 C 5.037361 0.000000
3 D 0.000000 3.394792
Since pandas 1.1.0 you can compute the sorting values with the key argument
df.sort_values('XP_home', key=lambda _: df[['XP_away', 'XP_home']].max(1), ascending=False)
Out:
item XP_home XP_away
1 B 6.412500 0.000000
0 A 0.000000 5.229861
2 C 5.037361 0.000000
3 D 0.000000 3.394792
My dataframe
class_lst = ["B","A","C","Z","H","K","O","W","L","R","M","Y","Q","X","X","G","G","G","G","G"]
value_lst = [1,0.999986,1,0.999358,0.999906,0.995292,0.998481,0.388307,0.99608,0.99829,1,0.087298,1,1,0.999993,1,1,1,1,1]
df =pd.DataFrame(
{'class': class_lst,
'val': value_lst
})
For any interval of 'val' in ranges
ranges = np.arange(0.0, 1.1, 0.1)
I would like to get the frequency of 'val' items, as follows:
class range frequency
A (0, 0.10] 0
A (0.10, 0.20] 0
A (0.20, 0.30] 0
...
A (0.90, 100] 1
G (0, 0.10] 0
G (0.10, 0.20] 0
G (0.20, 0.30] 0
...
G (0.80, 0.90] 0
G (0.90, 100] 5
...
I tried
df.groupby(pd.cut(df.val, ranges)).count()
but the output looks like
class val
val
(0, 0.1] 1 1
(0.1, 0.2] 0 0
(0.2, 0.3] 0 0
(0.3, 0.4] 1 1
(0.4, 0.5] 0 0
(0.5, 0.6] 0 0
(0.6, 0.7] 0 0
(0.7, 0.8] 0 0
(0.8, 0.9] 0 0
(0.9, 1] 18 18
and does not match with the expected one
This might be a good start:
df["range"] = pd.cut(df['val'], ranges)
class val range
0 B 1.000000 (0.9, 1.0]
1 A 0.999986 (0.9, 1.0]
2 C 1.000000 (0.9, 1.0]
3 Z 0.999358 (0.9, 1.0]
4 H 0.999906 (0.9, 1.0]
5 K 0.995292 (0.9, 1.0]
6 O 0.998481 (0.9, 1.0]
7 W 0.388307 (0.3, 0.4]
8 L 0.996080 (0.9, 1.0]
9 R 0.998290 (0.9, 1.0]
10 M 1.000000 (0.9, 1.0]
11 Y 0.087298 (0.0, 0.1]
12 Q 1.000000 (0.9, 1.0]
13 X 1.000000 (0.9, 1.0]
14 X 0.999993 (0.9, 1.0]
15 G 1.000000 (0.9, 1.0]
16 G 1.000000 (0.9, 1.0]
17 G 1.000000 (0.9, 1.0]
18 G 1.000000 (0.9, 1.0]
19 G 1.000000 (0.9, 1.0]
and then
df.groupby(["class", "range"]).size()
class range
A (0.9, 1.0] 1
B (0.9, 1.0] 1
C (0.9, 1.0] 1
G (0.9, 1.0] 5
H (0.9, 1.0] 1
K (0.9, 1.0] 1
L (0.9, 1.0] 1
M (0.9, 1.0] 1
O (0.9, 1.0] 1
Q (0.9, 1.0] 1
R (0.9, 1.0] 1
W (0.3, 0.4] 1
X (0.9, 1.0] 2
Y (0.0, 0.1] 1
Z (0.9, 1.0] 1
This will give already the right bin for each class and its frequency.
I have a DataFrame:
>>> df = pd.DataFrame({'row1' : [1,2,np.nan,4,5], 'row2' : [11,12,13,14,np.nan], 'row3':[22,22,23,24,25]}, index = 'a b c d e'.split()).T
>>> df
a b c d e
row1 1.0 2.0 NaN 4.0 5.0
row2 11.0 12.0 13.0 14.0 NaN
row3 22.0 22.0 23.0 24.0 25.0
and a Series that specifies the number of top N values I want from each row
>>> n_max = pd.Series([2,3,4])
What is Panda's way of using df and n_max to find the largest N elements of each (breaking ties with a random pick, just as .nlargest() would do)?
The desired output is
a b c d e
row1 NaN NaN NaN 4.0 5.0
row2 NaN 12.0 13.0 14.0 NaN
row3 22.0 NaN 23.0 24.0 25.0
I know how to do this with a uniform/fixed N across all rows (say, N=4). Note the tie-breaking in row3:
>>> df.stack().groupby(level=0).nlargest(4).unstack().reset_index(level=1, drop=True).reindex(columns=df.columns)
a b c d e
row1 1.0 2.0 NaN 4.0 5.0
row2 11.0 12.0 13.0 14.0 NaN
row3 22.0 NaN 23.0 24.0 25.0
But the goal, again, is to have row-specific N. Looping through each row obviously doesn't count (for performance reasons). And I've tried using .rank() with a mask but tie breaking doesn't work there...
Based on #ScottBoston's comment on the OP, it is possible to use the following mask based on rank to solve this problem:
>>> n_max.index = df.index
>>> df_rank = df.stack(dropna=False).groupby(level=0).rank(ascending=False, method='first').unstack()
>>> selected = df_rank.le(n_max, axis=0)
>>> df[selected]
a b c d e
row1 NaN NaN NaN 4.0 5.0
row2 NaN 12.0 13.0 14.0 NaN
row3 22.0 NaN 23.0 24.0 25.0
For performance, I would suggest NumPy -
def mask_variable_largest_per_row(df, n_max):
a = df.values
m,n = a.shape
nan_row_count = np.isnan(a).sum(1)
n_reset = n-n_max.values-nan_row_count
n_reset.clip(min=0, max=n-1, out = n_reset)
sidx = a.argsort(1)
mask = n_reset[:,None] > np.arange(n)
c = sidx[mask]
r = np.repeat(np.arange(m), n_reset)
a[r,c] = np.nan
return df
Sample run -
In [182]: df
Out[182]:
a b c d e
row1 1.0 2.0 NaN 4.0 5.0
row2 11.0 12.0 13.0 14.0 NaN
row3 22.0 22.0 5.0 24.0 25.0
In [183]: n_max = pd.Series([2,3,2])
In [184]: mask_variable_largest_per_row(df, n_max)
Out[184]:
a b c d e
row1 NaN NaN NaN 4.0 5.0
row2 NaN 12.0 13.0 14.0 NaN
row3 NaN NaN NaN 24.0 25.0
Further boost : Bringing in numpy.argpartition to replace the numpy.argsort should help, as we don't care about the order of indices to be reset as NaNs. Thus, a numpy.argpartition based one would be -
def mask_variable_largest_per_row_v2(df, n_max):
a = df.values
m,n = a.shape
nan_row_count = np.isnan(a).sum(1)
n_reset = n-n_max.values-nan_row_count
n_reset.clip(min=0, max=n-1, out = n_reset)
N = (n-n_max.values).max()
N = np.clip(N, a_min=0, a_max=n-1)
sidx = a.argpartition(N, axis=1) #sidx = a.argsort(1)
mask = n_reset[:,None] > np.arange(n)
c = sidx[mask]
r = np.repeat(np.arange(m), n_reset)
a[r,c] = np.nan
return df
Runtime test
Other approaches -
def pandas_rank_based(df, n_max):
n_max.index = df.index
df_rank = df.stack(dropna=False).groupby(level=0).rank\
(ascending=False, method='first').unstack()
selected = df_rank.le(n_max, axis=0)
return df[selected]
Verification and timings -
In [387]: arr = np.random.rand(1000,1000)
...: arr.ravel()[np.random.choice(arr.size, 10000, replace=0)] = np.nan
...: df1 = pd.DataFrame(arr)
...: df2 = df1.copy()
...: df3 = df1.copy()
...: n_max = pd.Series(np.random.randint(0,1000,(1000)))
...:
...: out1 = pandas_rank_based(df1, n_max)
...: out2 = mask_variable_largest_per_row(df2, n_max)
...: out3 = mask_variable_largest_per_row_v2(df3, n_max)
...: print np.nansum(out1-out2)==0 # Verify
...: print np.nansum(out1-out3)==0 # Verify
...:
True
True
In [388]: arr = np.random.rand(1000,1000)
...: arr.ravel()[np.random.choice(arr.size, 10000, replace=0)] = np.nan
...: df1 = pd.DataFrame(arr)
...: df2 = df1.copy()
...: df3 = df1.copy()
...: n_max = pd.Series(np.random.randint(0,1000,(1000)))
...:
In [389]: %timeit pandas_rank_based(df1, n_max)
1 loops, best of 3: 559 ms per loop
In [390]: %timeit mask_variable_largest_per_row(df2, n_max)
10 loops, best of 3: 34.1 ms per loop
In [391]: %timeit mask_variable_largest_per_row_v2(df3, n_max)
100 loops, best of 3: 5.92 ms per loop
Pretty good speedups there of 50x+ over the pandas built-in!
I'm trying to make a 3D scene with OpenGL ES 2, I'm new into xCode and Objective-C.
I follow this tutorial to transform blender *.obj generated file to *.h and *.c files
But, the script want obj like this:
v 1.000000 -1.000000 -1.000000
v 1.000000 -1.000000 1.000000
v -1.000000 -1.000000 1.000000
v -1.000000 -1.000000 -1.000000
v 1.000000 1.000000 -0.999999
v 0.999999 1.000000 1.000001
v -1.000000 1.000000 1.000000
v -1.000000 1.000000 -1.000000
vt 0.375624 0.500625
vt 0.624375 0.500624
vt 0.375625 0.749375
vt 0.375625 0.251875
vt 0.375624 0.003126
vt 0.624374 0.251874
vt 0.873126 0.749375
vt 0.873126 0.998126
vt 0.624375 0.749375
vt 0.624375 0.998126
vt 0.126874 0.998126
vt 0.126874 0.749375
vt 0.375625 0.998126
vt 0.624373 0.003126
vn 0.000000 -1.000000 0.000000
vn 0.000000 1.000000 0.000000
vn 1.000000 -0.000000 0.000000
vn 0.000000 -0.000000 1.000000
vn -1.000000 -0.000000 -0.000000
vn 0.000000 0.000000 -1.000000
vn 1.000000 0.000000 0.000001
s off
f 1/1/1 2/2/1 4/3/1
f 5/4/2 8/5/2 6/6/2
f 1/1/3 5/4/3 2/2/3
f 2/7/4 6/8/4 3/9/4
f 3/9/5 7/10/5 4/3/5
f 5/11/6 1/12/6 8/13/6
f 2/2/1 3/9/1 4/3/1
f 8/5/2 7/14/2 6/6/2
f 5/4/7 6/6/7 2/2/7
f 6/8/4 7/10/4 3/9/4
f 7/10/5 8/13/5 4/3/5
f 1/12/6 4/3/6 8/13/6
And when I create a new cube (or anything else) I obtain an obj like this:
v 1.000000 -1.000000 -1.000000
v 1.000000 -1.000000 1.000000
v -1.000000 -1.000000 1.000000
v -1.000000 -1.000000 -1.000000
v 1.000000 1.000000 -0.999999
v 0.999999 1.000000 1.000001
v -1.000000 1.000000 1.000000
v -1.000000 1.000000 -1.000000
vn 0.000000 -1.000000 0.000000
vn 0.000000 1.000000 0.000000
vn 1.000000 -0.000000 0.000000
vn 0.000000 -0.000000 1.000000
vn -1.000000 -0.000000 -0.000000
vn 0.000000 0.000000 -1.000000
vn 1.000000 0.000000 0.000001
s off
f 1//1 2//1 4//1
f 5//2 8//2 6//2
f 1//3 5//3 2//3
f 2//4 6//4 3//4
f 3//5 7//5 4//5
f 5//6 1//6 8//6
f 2//1 3//1 4//1
f 8//2 7//2 6//2
f 5//7 6//7 2//7
f 6//4 7//4 3//4
f 7//5 8//5 4//5
f 1//6 4//6 8//6
Faces are separated by two slashes where I expected only one.
There is any simpliest way to generate .h and .c files for xCode, every script (for Blender) I try failed. Or anyone can tell me how to get a clean obj file.
Thanks a lot
Your resulting file is correct obj. However, what you want it to be - is incorrect obj.
Format for f command is f position_id/texture_coordinates_id/normal_id. You don't have texture coordinates, so this field is empty.
Options are
fix parser so it could load obj without texture coordinates, or
just add UV map to your object.