how to break a for loop?

[Gregory Petrosyan]

Hello!
It's 1:56 o'clock in St.-Petersburg now, and I am still coding... maybe
that's why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can't :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

but it removes ALL zeros from list. What's up?
P.S. I feel SO stupid asking this quastion... ;-)

[Schüle Daniel]

Gregory Petrosyan wrote:

Hello!
It's 1:56 o'clock in St.-Petersburg now, and I am still coding... maybe
that's why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can't :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

but it removes ALL zeros from list. What's up?
I don't know how enumerate is implemented, but I would
suspect that modifying the list object in the loop
through del is asking for trouble

try
for i,coef in enumerate(coefs[:]):
instead
P.S. I feel SO stupid asking this quastion... ;-)

uda4i

hth, Daniel

[Petr Jakes]

zero_list=[0,0,0,0,0,1,2,3,4]
for x in range(len(zero_list)):
if zero_list[x]!=0: break
nonzero_list=zero_list[x:]
print nonzero_list

but some more "pythonic" solutions will be posted from other users I
guess :)

HTH
Petr Jakes

[Schüle Daniel]

I just spend some more seconds thinking about it :)

lst = [0,0,0,1,2,3,0,11]

try:
del lst[0:lst.index(0)]
except ValueError:
pass

is better solution

[Jesus Rivero - (Neurogeek)]

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Hello Gregory!

I would also use lists to implement this. They are more practical and
you could take advantage of pop() or remove() methods instead of using
the Oh-so-Scary del. You could also use a lambda+map or reduce methods
to remove all zeros or just do something like this:

list_ = [0,0,0,1,0,0,1]
print [ elm for elm in list_ if elm != 0 ]
[1,1]

Regards,

Jesus (Neurogeek)

Gregory Petrosyan wrote:

Hello!
It's 1:56 o'clock in St.-Petersburg now, and I am still coding... maybe
that's why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can't :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

but it removes ALL zeros from list. What's up?
P.S. I feel SO stupid asking this quastion... ;-)

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[Scott David Daniels]

Gregory Petrosyan wrote:

Hello!
It's 1:56 o'clock in St.-Petersburg now, and I am still coding... maybe
that's why I encountered stupid problem: I need to remove zeros from
the begining of list, but I can't :-(. I use

for i,coef in enumerate(coefs):
if coef == 0:
del coefs[i]
else:
break

for index, coef in enumerate(coefs):
if coef:
if index:
del coefs[: index]
break

--Scott David Daniels
sc***********@acm.org

[bonono]

Gregory Petrosyan wrote:

I need to remove zeros from
the begining of list, but I can't :-(.

I believe the following is almost a direct translation of the above
sentence.

import itertools as it
a=[0,0,0,1,0]
a[:]=it.dropwhile(lambda x: x is 0, a)

[Steven D'Aprano]

Petr Jakes wrote:

zero_list=[0,0,0,0,0,1,2,3,4]
for x in range(len(zero_list)):
if zero_list[x]!=0: break
nonzero_list=zero_list[x:]
print nonzero_list

but some more "pythonic" solutions will be posted from other users I
guess :)

That looks perfectly Pythonic to me.

You know, just because Python has for loops and
multi-line blocks of code doesn't mean we shouldn't use
them *wink*
--
Steven.

[Giovanni Bajo]

bo****@gmail.com wrote:

I need to remove zeros from
the begining of list, but I can't :-(.

I believe the following is almost a direct translation of the above
sentence.

import itertools as it
a=[0,0,0,1,0]
a[:]=it.dropwhile(lambda x: x is 0, a)

Usa lambda x: x==0 or simply lambda x: x

Using 'is' relies on the interpreter reusing existing instances of the
immutable object '0', which is an implementation detail.
--
Giovanni Bajo

[bonono]

Giovanni Bajo wrote:

bo****@gmail.com wrote:

I need to remove zeros from
the begining of list, but I can't :-(.

I believe the following is almost a direct translation of the above
sentence.

import itertools as it
a=[0,0,0,1,0]
a[:]=it.dropwhile(lambda x: x is 0, a)

Usa lambda x: x==0 or simply lambda x: x

Using 'is' relies on the interpreter reusing existing instances of the
immutable object '0', which is an implementation detail.

stand corrected. But I don't think "lambda x: x" is I want as 0 would
be false which would stop the dropwhile right away.

For this particular case one can even use operator.not_ directly, which
skip the lambda(has some peformance advantage).

[Gregory Petrosyan]

Thanks to everybody for help. Python community is very friendly and
helpfull indeed!

[bearophileHUGS]

This time it seems that using itertools gives slower results, this is
the test code:

import itertools as it
from operator import __not__

def trim_leading_zeros(seq):
if not seq:
return seq
for pos, el in enumerate(seq):
if el != 0:
break
if seq[pos] == 0:
del seq[:]
return seq
return seq[pos:]

def trim_leading_zeros2(seq):
seq[:] = it.dropwhile(__not__, seq)
return seq

data = ([0,0,0,0,0,1,2,3,4],
[1,2,3,4],
[0,1],
[0,0,0],
[0,0,0,1],
[])

for l in data:
print l, trim_leading_zeros(l), trim_leading_zeros2(l)

from time import clock
n = 3 * 10**5

l = [0] * n + [1] * n
t = clock()
trim_leading_zeros(l)
print round(clock()-t, 2), "s"

l = [0] * n + [1] * n
t = clock()
trim_leading_zeros2(l)
print round(clock()-t, 2), "s"

Gives this output on my PC:

[0, 0, 0, 0, 0, 1, 2, 3, 4] [1, 2, 3, 4] [1, 2, 3, 4]
[1, 2, 3, 4] [1, 2, 3, 4] [1, 2, 3, 4]
[0, 1] [1] [1]
[0, 0, 0] [] []
[0, 0, 0, 1] [1] [1]
[] [] []
0.3 s
2.86 s

Bye,
bearophile

[bonono]

[e-mail] wrote:

This time it seems that using itertools gives slower results, this is
the test code:

Understandable, as dropwhile still needs to go through the whole list
and the difference will be larger when the list get longer. Though I
still prefer that if the list is not horribly long as it is like a
spec. But then, one can always name the fast in place del slice version
like "dropwhileFast".