filling an array with a loop in python

Question

my problem is as follows, i have an array named crave=['word1','word2','word3'....'wordx'] and i want to transform into ITEM=[['word1','word2','word3'],['word4','word5','word6'] etc] i used the following code

buff=[['none','none','none']]
n=10
ITEM=[]
i=0
while 1>0 :


 buff[0][0]=crave[i]
 buff[0][1]=crave[i+1]
 buff[0][2]=crave[i+2]
 ITEM.insert(len(ITEM),buff[0])
 i=i+3
 if i>n:
      break

but what i get instead is [['wordx-2','wordx-1','wordx'],['wordx-2','wordx-1','wordx'],['wordx-2','wordx-1','wordx']etc] why does this happen ?:(

so you want to split the list into 3 elements sublists? is ther ordering important? — Moj
– Moj, Commented May 26, 2015 at 12:58
I tried to understand what your code does, I had to change some lines as to get it compile and run. Here's what I made of it: ideone.com/bHuDx2 — Wolf
– Wolf, Commented May 26, 2015 at 13:48

ZdaR · Accepted Answer · 2015-05-26 12:57:58Z

10

You can easily do this by using list comprehension. xrange(0, len(crave), 3) is used to iterate from 0 to len(crave) with an interval of 3. and then we use list slicing crave[i:i+3] to extract the required data.

crave=['word1','word2','word3','word4','word5','word6','word7']

ITEM = [crave[i:i+3] for i in xrange(0, len(crave), 3)]

print ITEM
>>> [['word1', 'word2', 'word3'], ['word4', 'word5', 'word6'], ['word7']]

answered May 26, 2015 at 12:57

ZdaR

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2 Comments

Wolf Over a year ago

Interesting, especially the processing of the last ("broken") triple - it really works :-) ideone.com/S9AQF7

ZdaR Over a year ago

Thanks for appreciation @Wolf

Haresh Shyara · Accepted Answer · 2015-05-26 13:31:52Z

4

try to this.

crave = ['word1', 'word2', 'word3', 'word4', 'word5', 'word6', 'word7']
li = []
for x in range(0, len(crave), 3):
    li.append(crave[x:x+3])
print li

answered May 26, 2015 at 13:31

Haresh Shyara

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1 Comment

Wolf Over a year ago

...list comprehension explained ;-)

Moj · Accepted Answer · 2015-05-26 13:03:01Z

1

import numpy as np
In [10]: a =np.arange(20)

In [11]: a
Out[11]: 
array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16,
       17, 18, 19])

In [12]: np.array_split(a,len(a)/3)
Out[12]: 

[array([0, 1, 2, 3]),
 array([4, 5, 6, 7]),
 array([ 8,  9, 10]),
 array([11, 12, 13]),
 array([14, 15, 16]),
 array([17, 18, 19])]

answered May 26, 2015 at 13:03

Moj

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Comments

Rob · Accepted Answer · 2015-05-26 14:02:19Z

1

In respect to the question of "why does this happen":

The problem is that buff[0] is being referenced by the .insert, not copied. Hence on the next turn of the loop when buff changes anything that references buff will also change.

To do an explicit copy use list(). E.g.

 ITEM.insert(len(ITEM),list(buff[0]))

edited May 26, 2015 at 14:02

answered May 26, 2015 at 13:49

Rob

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