我正在尝试从Matlab迁移到Python。我正在将Matlab中的一些代码重写为Python以进行测试。我已经安装了Anaconda，目前正在使用Spyder IDE。使用Matlab，我创建了一个函数，该函数返回更接近函数输入参数的商用API 5L直径（直径）和管道厚度（espesor）的值。我是使用Matlab表完成此操作的。
请注意，直径（diametro_entrada）和厚度（espesor_entrada）的输入单位为米[m]，函数内部的厚度单位为毫米[mm]，这就是为什么最后我必须将espesor_entrada * 1000 < / p>
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
使用输入直径（d）和输入厚度（e），我得到直径小于0.05且厚度小于前者1.2的商用管。
我想用Numpy或其他软件包在Python中重现它。 首先，我定义了2个Numpy数组，其名称与Matlab中的名称相同，但用逗号分隔而不是分号，并且每行末尾都没有“ ...”，然后将另一个Numpy数组定义为：
TablaAPI=numpy.array([Diametro_m,Espesor_mm])
我想知道我是否可以像在Matlab中那样以某种方式索引该数组，或者必须定义完全不同的其他东西。
非常感谢！
答案 0 :(得分：1)
您肯定可以！
以下是如何使用numpy的示例：
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
如果您需要执行更重的查询或混合列数据类型，数据框也可能对您的应用程序非常有用。如果您选择该选项，则此代码将为您工作：
# 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)
答案 1 :(得分：0)
由于您没有提供预期输出的示例，因此您可能会猜测您的实际需求，但这是一个带有numpy的版本。
# 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)
或者您可以只用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)))
一旦您使用python，就有无数种方法可以执行此操作，您可以轻松地对pandas或sqlite进行相同的操作。我个人的喜好倾向于尽可能少地依赖，在这种情况下，我会选择一个csv文件作为输入，然后在不使用numpy的情况下进行处理，如果这是一个真正的大规模问题，我会考虑使用sqlite / numpy / pandas。 / p>
祝您过渡顺利，我认为您不会后悔。