Azimuth
Reputation: 45
Index by condition in python-numpy?
I'm trying to migrate from Matlab to Python. I'm rewriting some code that I had in Matlab to Python for testing. I've installed Anaconda and currently using Spyder IDE. Using Matlab I created a function that returns the values of the commercial API 5L diameter(diametro) and thickness(espesor) of pipes that are closer to the input parameters of the function. I did this using a Matlab table.
Note that the inputs of the diameter(diametro_entrada) and thickness(espesor_entrada) are in meters[m] and the thickness inside the function are in millimeters [mm], that's why at the end I had to multiply espesor_entrada*1000
function tabla_seleccion=tablaAPI(diametro_entrada,espesor_entrada)
%Proporciona la tabla de caños API 5L, introducir diámetro en [m] y espesor
%en [m]
Diametro_m=[0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;0.3556;...
0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;0.4064;...
0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;0.4570;...
0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;0.5080;...
0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;0.559;...
0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;0.610;...
0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;0.660;...
0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;0.711;...
0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;0.762;...
0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813;0.813];
Espesor_mm=[4.8;5.2;5.3;5.6;6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;31.8;...
4.8;5.2;5.6;6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;...
4.8;5.6;6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;...
5.6;6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;33.3;34.9;...
5.6;6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;33.3;34.9;36.5;38.1;...
6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;33.3;34.9;36.5;38.1;39.7;...
6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;...
6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;...
6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8;...
6.4;7.1;7.9;8.7;9.5;10.3;11.1;11.9;12.7;14.3;15.9;17.5;19.1;20.6;22.2;23.8;25.4;27.0;28.6;30.2;31.8];
TablaAPI=table(Diametro_m,Espesor_mm);
tabla_seleccion=TablaAPI(abs(TablaAPI.Diametro_m-diametro_entrada)<0.05 & abs(TablaAPI.Espesor_mm-(espesor_entrada*1000))<1.2,:);
end
With the input diameter(d) and the input thickness(e) I get the commercial pipe that has less than 0.05 in diameter and 1.2 in thickness from the former.
I want to do reproduce this in Python with Numpy or another package. First I defined 2 Numpy arrays, with the same names as in Matlab but comma separated instead of semicolon and without the "..." at the end of each line, then defined another Numpy array as:
TablaAPI=numpy.array([Diametro_m,Espesor_mm])
I want to know if I can index that array in some way like I did in Matlab or I have to define something else totally different.
Thanks a lot!
Upvotes: 0
Views: 120
Answers (2)
ahed87
Reputation: 1360
Since you have not given an example of your expected output it's a bit of guessing what you are really after, but here is one version with numpy.
# rewritten arrays for numpy
Diametro_m=[0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,0.3556,
0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,0.4064,
0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,0.4570,
0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,0.5080,
0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,0.559,
0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,0.610,
0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,0.660,
0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,0.711,
0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,0.762,
0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813,0.813]
Espesor_mm=[4.8,5.2,5.3,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,31.8,
4.8,5.2,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,
4.8,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,
5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9,
5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9,36.5,38.1,
6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9,36.5,38.1,39.7,
6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,
6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,
6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,
6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8]
import numpy as np
diametro_entrada = 0.4
espesor_entrada = 5
Diametro_m = np.array(Diametro_m)
Espesor_mm = np.array(Espesor_mm)
# Diametro_m and Espesor_mm has shape (223,)
# if not change so that they have that shape
table = np.array([Diametro_m, Espesor_mm]).T
mask = np.where((np.abs(Diametro_m - diametro_entrada) < 0.05) &
(np.abs(Espesor_mm - espesor_entrada) < 1.2)
)
result = table[mask]
print('with numpy')
print(result)
or you can do it with just python...
# redo with python only
# based on a simple dict and list comprehension
D_m = [0.3556, 0.4064, 0.4570, 0.5080, 0.559, 0.610, 0.660, 0.711, 0.762, 0.813]
E_mm = [[4.8,5.2,5.3,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,31.8],
[4.8,5.2,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8],
[4.8,5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8],
[5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9],
[5.6,6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9,36.5,38.1],
[6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8,33.3,34.9,36.5,38.1,39.7],
[6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4],
[6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4],
[6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8],
[6.4,7.1,7.9,8.7,9.5,10.3,11.1,11.9,12.7,14.3,15.9,17.5,19.1,20.6,22.2,23.8,25.4,27.0,28.6,30.2,31.8]]
table2 = dict(zip(D_m, E_mm))
result2 = []
for D, E in table2.items():
if abs(D - diametro_entrada) < 0.05:
Et = [t for t in E if abs(t - espesor_entrada) < 1.2]
result2 += [(D, t) for t in Et]
print('with vanilla python')
print('\n'.join((str(r) for r in result2)))
Once you are in python there are endless ways to do this, you could easily do the same with pandas, or sqlite. My personal preference tends to lean towards as little dependencies as possible, in this case I would go for a csv file as input and then do it without numpy, if it was a true large scale problem I would consider sqlite/numpy/pandas.
Good luck with the transition, I don't think you will regret it.
Upvotes: 0
Daniel Scott
Reputation: 985
You sure can!
Here's an example of how you can use numpy:
Using Numpy
import math
import numpy as np
# Declare your Diametro_m, Espesor_mmhere just like you did in your example
# Transpose and merge the columns
arr = np.concatenate((Diametro_m, Espesor_mm.T), axis=1)
selection = arr[np.ix_(abs(arr[:0])<0.05,abs(arr[:1]-(math.e*1000)) > <1.2 )]
Example usage from John Zwinck's answer
Using Dataframes
Dataframes may also be great for your application in case you need to do heavier queries or mix column datatypes. This code should work for you, if you choose that option:
# These imports go at the top of your document
import pandas as pd
import numpy as np
import math
# Declare your Diametro_m, Espesor_mmhere just like you did in your example
df_d = pd.DataFrame(data=Diametro_m,
index=np.array(range(1, len(Diametro_m))),
columns=np.array(range(1, len(Diametro_m))))
df_e = pd.DataFrame(data=Espesor_mm,
index=np.array(range(1, len(Diametro_m))),
columns=np.array(range(1, len(Diametro_m))))
# Merge the dataframes
merged_df = pd.merge(left=df_d , left_index=True
right=df_e , right_index=True,
how='inner')
# Now you can perform your selections like this:
selection = merged_df.loc[abs(merged_df['df_d']) <0.05, abs(merged_df['df_e']-(math.e*1000))) <1.2]
# This "mask" of the dataframe will return all results that satisfy your query.
print(selection)
Upvotes: 1
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