在python中用循环填充数组

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 ?:(

Answer 1

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']]

Answer 2

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

Answer 3

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])]

Answer 4

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(). Eg

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