dictionaries create table python

Question

I don't get it why here the result is [[0 0] [0 0]]; Here is the code:

def create_table(m,n):
    t=[]
    one_line=[]
    for i in range(0,m):
        one_line.append(0)
    for i in range(0,n):
        t.append(one_line)
    return t

print(create_table(2,2));

After first iteration in one_line we have [0] and in t [[0]].

After the second iteration, where i=2 we have one_line = [0 0] and in t = [[0 0] [0 0]]. But why is not t=[[0] [0 0]] because the previous t was [[0]] not [[0 0] [0 0]].

Any explanation?

I don't understand it………what do you mean by "the previous t"? — laike9m
– laike9m, Commented Apr 14, 2016 at 11:16
we have the first t from the first iteration where it was [[0]]. — Irina Farcau
– Irina Farcau, Commented Apr 14, 2016 at 11:18
i think, you have understood python loop false. the second loop will run after the the first loop is finished. the previous t of your code was [[0, 0]] and not [[0]] — qvpham
– qvpham, Commented Apr 14, 2016 at 11:25

BartDur · Accepted Answer · 2016-04-14 11:20:22Z

1

You have defined two for loops:

First you loop twice through the first for loop, creating one_line=[0,0]
Then you loop trough the second loop (twice), thus appending [0,0] twice to t thus ending up with t=[[0,0],[0,0]]

answered Apr 14, 2016 at 11:20

BartDur

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qvpham · Accepted Answer · 2016-04-14 11:33:56Z

0

So run your function call create_table(2, 2)

create_table(2, 2):

set: m = 2, n = 2

t = []
one_line = []

define: t = [], one_line = []

for i in range(0,m):

for i in [0, 1]

with i = 0

one_line.append(0)

one_line = [0]

with i = 1

one_line.append(0)

one_line = [0, 0]

for i in range(0,n):

for i in (0, 1)

with i = 0

t.append(one_line)

t = [[0, 0]]

with i = 1

t.append(one_line)

t = [[0, 0], [0, 0]]

answered Apr 14, 2016 at 11:33

qvpham

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