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 2 years ago
There is this issue about lists in python I have encountered which is probably about the code I wrote so some insight will be appreciated.
Problem Wk.6.1.3: A bigger bunch of zeros and code goes wrong is :
def zeroVector(n):
a=list(range(n))
b=[]
k=0
while k<n:
b.append(0)
k+=1
for e in a:
a[e]=0
print(a)
return a
def zeroArray(m,n):
foo=zeroVector(m)
print(foo)
for f in foo:
foo[f]=zeroVector(n)
print(foo)
return foo
which has this return for zeroArray(3,4):
[[0, 0, 0, 0], 0, 0]
 2 years ago
There is this issue about lists in python I have encountered which is probably about the code I wrote so some insight will be appreciated. Problem Wk.6.1.3: A bigger bunch of zeros and code goes wrong is : def zeroVector(n): a=list(range(n)) b=[] k=0 while k<n: b.append(0) k+=1 for e in a: a[e]=0 print(a) return a def zeroArray(m,n): foo=zeroVector(m) print(foo) for f in foo: foo[f]=zeroVector(n) print(foo) return foo which has this return for zeroArray(3,4): [[0, 0, 0, 0], 0, 0]

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rapidsnow
 2 years ago
Best ResponseYou've already chosen the best response.1Are you trying to generate a two dimensional list of zeros? I'm not sure I understand your purpose, but if so there are a few ways to do it. For a list: def zeroArray(m, n): return [[0]*n]*m Alternatively, if you want a numpy array: import numpy as np def zeroArray(m,n): return np.zeros(m*n).reshape(m,n)

rapidsnow
 2 years ago
Best ResponseYou've already chosen the best response.1Also, I see what you did wrong in your code, if you do want to do it that way. You say: foo=zeroVector(m) print(foo) for f in foo: foo[f]=zeroVector(n) print(foo) However, look at what foo is. It is zeroVector(m), which outputs a vector full of zeros that is of length m. So you are looping over values of zero m times. Essentially you are doing foo[0]=[0,0....,0] a lot of times. Your for statement should be 'for f in range(len(foo)):'

Sarper
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you a lot.Apart from the first statement does it nicely, now I know a bit more about what numpy does that I didnt bother when it refused to install (have both python 2.7 and 3.3).However from my ambigious question this remains still; if foo is a list and "for ... in" statement iterates on any list's elements why should use range(len(foo)). And once again you've earned best credits for your great help already

rapidsnow
 2 years ago
Best ResponseYou've already chosen the best response.1Imagine a list ">>> alphabet = ['a', 'b', 'c']" When you do for x in alphabet: print x What you are doing is iterating over the elements in alphabet. So the output would be: 'a' 'b' 'c' So in the original statement, you say ">>>for f in foo:". But foo is a list of zeros. So you are repeatedly giving f a value of zero. You then loop through foo[f]. So what are you doing? You are assigning things to the 0th index of foo and then assigning something else to foo[0]. You are saying foo[0] = [0,0,0...,0] a lot of times in a row. range(n) creates a list object that goes from 0 > n1. [0, 1, 2, 3, ..., n1]. You are trying to access the f index of foo. Now you are saying foo[0] = [0,0,0...,0] foo[1] = [0,0,0...,0] ... foo[n1] = [0,0,0...,0] Which is what you were trying to do.

Sarper
 2 years ago
Best ResponseYou've already chosen the best response.0So if I understand correctly iteration element f (f in foo) has the value of 0 (more elaborately any item's value by an index number from 0 to top which is not shown).I got it now.I began to dig through some other lessons as well and YOU have been a great help. Yet when everything looked settled a new issue arose;if we use the code below def zeroArray(m, n): return [[0]*n]*m and try to assign 2nd element of the 2nd element of constructed list say like: z=zeroArray(3,4) by z[2][2]=1 value of z becomes; [[0, 0, 1, 0], [0, 0, 1, 0], [0, 0, 1, 0]] you are already a charm even if you cant tell me if any way exists to make such assignment
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