python创建数组并运行,在for循环中创建新数组（Python）

左手的ㄟ右手 2022-10-04 13:00 209阅读 0赞

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I'm preparing a data set to run in the program rpy (R, which runs in Python) for statistical analysis. It looks like this:

data = \[\[0, 1, 0, 0, 0, 0, 0, 1, 0, 0\], \[0, 0, 1, 0, 0, 1, 0, 0, 0, 0\],

\[0, 1, 1, 0, 0, 0, 0, 0, 0, 1\], \[1, 0, 0, 0, 1, 0, 0, 0, 0, 1\],

\[0, 0, 1, 1, , 0, 0, 0, 0, 0\], \[0, 0, 0, 0, 1, 1, 0, 0, 0, 0\],

\[1, 0, 0, 0, 0, 0, 0, 1, 0, 0\], \[0, 1, 0, 0, 0, 1, 0, 0, 0, 0\],

\[0, 0, 0, 0, 1, 0, 0, 0, 1, 0\]\]

For me to use this data, I need to isolate the dependent variable (y) from the independent ones (x). I need to create a new list for each column for year as such:

y = data\[:,9\]

x1 = data\[:,0\]

x2 = data\[:,1\]

x3 = data\[:,2\]

x4 = data\[:,3\]

x5 = data\[:,4\]

x6 = data\[:,5\]

x7 = data\[:,6\]

x8 = data\[:,7\]

x9 = data\[:,8\]

x10 = data\[:,9\]

Suppose my data has 67 columns. Is there a way to loop through all the columns and create each one automatically without having to type out all of them? I do not want to hard code all the arrays up to 67.

Something along the lines of this, but it doesn't work:

i=0

for d in data:

"x%d"%i = data\[:,i-1\]

i+=1

This is the rest of the code:

rpy.set\_default\_mode(rpy.NO\_CONVERSION)

linear\_model = rpy.r.lm(rpy.r("y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10"), data = rpy.r.data\_frame(x1=x1,x2=x2,x3=x3,x4=x4,x5=x5,x6=x6,x7=x7,x8=x8,x9=x9,x10=x10,y=y))

rpy.set\_default\_mode(rpy.BASIC\_CONVERSION)

print linear\_model.as\_py()\['coefficients'\]

summary = rpy.r.summary(linear\_model)

解决方案

Why not try something like this to transpose the columns:

x = \[\]

for d in xrange(0,66):

x.append(data\[:,d\])

Unless it's absolutely essential that there is a separate data structure for each item, although I don't know why you would need separate data strucures...

EDIT: If not here's something that should work precisely the way you described:

for d in xrange(1,68):

exec 'x%s = data\[:,%s\]' %(d,d-1)

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