Is there any efficient way to define N variables and then use each variable to define one list using python?

Question

z1=z2=z3=z4=z5=z6=[]

for i2 in range(0,len(a),L):

    z1.append(a[i2])
print 'part 1:'

print ''.join(z1)

for i2 in range(1,len(a),L):

    z2.append(a[i2])

print 'part 2:'


print ''.join(z2)

for i2 in range(2,len(a),L):

    z3.append(a[i2])

print 'part 3:'

print ''.join(z3)

for i2 in range(3,len(a),L):

    z4.append(a[i2])

print 'part 4:'

print ''.join(z4)

for i2 in range(4,len(a),L):

    z5.append(a[i2])

print 'part 5:'

print ''.join(z5)

for i2 in range(5,len(a),L):

    z6.append(a[i2])

print 'part 6:'

print ''.join(z6)

I want to define lists up to z152. But I don't know, how can I store 152 different data array without defining (z1 to z152) ? Is there any efficient way to this ?

After that I want to retrieve each list as per my requirement.

z=list(xrange(0,len(a)))
v1=0
#L will be as per program requirement
while v1<L:
    for i2 in xrange(v1,len(a),L):
        z[v1].append(a[i2])
    print ''.join(z[v1])
    print len(z[v1])
    v1+=1

I tried to run above code but it's shown up with below error.

AttributeError: 'int' object has no attribute 'append'.

So how can I filter this error ?

This is not the way to solve your problem. Use a parent list (say, `z`) that holds the 152 inner lists and access to them using `z[0], z[1], ... , z[151]` instead. — Selcuk, Oct 07 '16 at 00:43
Also note that `z1` through `z6` all point to the same list. Changing one of them will change the rest, because there's only one list. — TigerhawkT3, Oct 07 '16 at 00:46

Is there any efficient way to define N variables and then use each variable to define one list using python?

0 Answers0