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Join Date: Oct 2007

Posts: 7

Reputation: sjvr767 is an unknown quantity at this point

Solved Threads: 1

sjvr767

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Newbie Poster

comparing lists

Aug 26th, 2008

Does any one know of any modules which could help me in comparing lists.

I have the following problem. There are lists with names and there are lists with associated weights. For example,

a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]

b = ['a', 'c']
w_b = [0.5, 0.5]

I need to write a function that will compare a and b and insert 0s into the appropriate positions within w_b.... i.e. the output should be a list [0.5, 0, 0.5, 0].

I have tried the following, but it goes into an endless loop!

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
def comp_lists(list1, list2):
	length = max(len(list1), len(list2))
	i = 0
	while i < length:
		a = list1[i]
		b = list2[i]
		if a != b and min(a, b) != 0:
			break
		else:
			i = i + 1
	return i
 
def insert_zero_weight(share_list1, share_list2, w1, w2):
	W1 = list(w1)
	W2 = list(w2)
	if share_list1 == share_list2:
		return "List are equal"
	i = comp_lists(share_list1, share_list2)
	print i
	print max(len(share_list1), len(share_list2))
	print len(share_list1), len(share_list2)
	if i < max(len(share_list1), len(share_list2)) and len(share_list1) < len(share_list2):
		W1.insert(i, 0)
		return W1
	elif i < max(len(share_list1), len(share_list2)) and len(share_list1) > len(share_list2):
		W2.insert(i, 0)
		return W2
	else:
		return "Error"
 
def Insert_zero_weight(share_list1, share_list2, w1, w2):
	W1 = list(w1)
	W2 = list(w2)
	if W1 < W2:
		while W1 < W2:
			print 'in is %d long' %len(W1)
			W3 = insert_zero_weight(share_list1, share_list2, W1, W2)
			print 'out is %d long' %len(W3)
			del W1
			W1 = list(W3)
		return W1
	elif W1 > W2:
		while W2 < W1:
			print 'in is %d long' %len(W2)
			W3 = insert_zero_weight(share_list1, share_list2, W1, W2)
			print 'out is %d long' %len(W3)
			del W2
			W2 = list(W3)
		return W2
	else:
		return "Error"
def comp_lists(list1, list2):
	length = max(len(list1), len(list2))
	i = 0
	while i < length:
		a = list1[i]
		b = list2[i]
		if a != b and min(a, b) != 0:
			break
		else:
			i = i + 1
	return i
	
def insert_zero_weight(share_list1, share_list2, w1, w2):
	W1 = list(w1)
	W2 = list(w2)
	if share_list1 == share_list2:
		return "List are equal"
	i = comp_lists(share_list1, share_list2)
	print i
	print max(len(share_list1), len(share_list2))
	print len(share_list1), len(share_list2)
	if i < max(len(share_list1), len(share_list2)) and len(share_list1) < len(share_list2):
		W1.insert(i, 0)
		return W1
	elif i < max(len(share_list1), len(share_list2)) and len(share_list1) > len(share_list2):
		W2.insert(i, 0)
		return W2
	else:
		return "Error"

def Insert_zero_weight(share_list1, share_list2, w1, w2):
	W1 = list(w1)
	W2 = list(w2)
	if W1 < W2:
		while W1 < W2:
			print 'in is %d long' %len(W1)
			W3 = insert_zero_weight(share_list1, share_list2, W1, W2)
			print 'out is %d long' %len(W3)
			del W1
			W1 = list(W3)
		return W1
	elif W1 > W2:
		while W2 < W1:
			print 'in is %d long' %len(W2)
			W3 = insert_zero_weight(share_list1, share_list2, W1, W2)
			print 'out is %d long' %len(W3)
			del W2
			W2 = list(W3)
		return W2
	else:
		return "Error"

Where you would run

Help with Code Tags

Python Syntax (Toggle Plain Text)

Insert_zero_weight(a, b, w_a, w_b)

.

Any help, whatsoever, would be appreciated.

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Join Date: Jul 2005

Posts: 1,221

Reputation: bumsfeld will become famous soon enough

Solved Threads: 137

bumsfeld

Offline

Nearly a Posting Virtuoso

Re: comparing lists

Aug 26th, 2008

At first blush i would say this might do:

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
# compare these two list set and convert
# w_b into [0.5, 0, 0.5, 0]
 
a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]
 
b = ['a', 'c']
w_b = [0.5, 0.5]
 
 
for ix, aa in enumerate(a):
    if aa not in b:
        w_b.insert(ix, 0)
 
print w_b  # result --> [0.5, 0, 0.5, 0]
# compare these two list set and convert
# w_b into [0.5, 0, 0.5, 0]

a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]

b = ['a', 'c']
w_b = [0.5, 0.5]


for ix, aa in enumerate(a):
    if aa not in b:
        w_b.insert(ix, 0)
        
print w_b  # result --> [0.5, 0, 0.5, 0]

Also note that you can combine two related lists in one container:

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
# combine two lists to show their relationship better
 
a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]
 
# list of (name, weight) tuples
print zip(a, w_a)  # [('a', 0.25), ('b', 0.25), ('c', 0.25), ('d', 0.25)]
 
# dictioary with name:weight pairs
print dict(zip(a, w_a)) # {'a': 0.25, 'c': 0.25, 'b': 0.25, 'd': 0.25}
# combine two lists to show their relationship better

a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]

# list of (name, weight) tuples
print zip(a, w_a)  # [('a', 0.25), ('b', 0.25), ('c', 0.25), ('d', 0.25)]

# dictioary with name:weight pairs
print dict(zip(a, w_a)) # {'a': 0.25, 'c': 0.25, 'b': 0.25, 'd': 0.25}

BTW, dictionaries are much faster way to look things up then indexing two separate related lists:

 Help with Code Tags 
 python Syntax (Toggle Plain Text) 
a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]
 
b = ['a', 'c']
w_b = [0.5, 0.5]
 
dict_a = dict(zip(a, w_a))
dict_b = dict(zip(b, w_b))
 
for k, v in dict_a.items():
    if k not in dict_b.keys():
        dict_b[k] = 0
 
print dict_b  # {'a': 0.5, 'c': 0.5, 'b': 0, 'd': 0}
a = ['a', 'b', 'c', 'd']
w_a = [0.25, 0.25, 0.25, 0.25]

b = ['a', 'c']
w_b = [0.5, 0.5]

dict_a = dict(zip(a, w_a))
dict_b = dict(zip(b, w_b))

for k, v in dict_a.items():
    if k not in dict_b.keys():
        dict_b[k] = 0

print dict_b  # {'a': 0.5, 'c': 0.5, 'b': 0, 'd': 0}

Last edited by bumsfeld; Aug 26th, 2008 at 11:56 am.

Should you find Irony, you can keep her!

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Join Date: Oct 2007

Posts: 7

Reputation: sjvr767 is an unknown quantity at this point

Solved Threads: 1

sjvr767

Offline

Newbie Poster

Re: comparing lists

Aug 27th, 2008

Thank you! This works so well...

I am using your first snippet of code, the one using enumerate. While I realise that dictionary would be faster, lists fit in better with how the rest of the script is programmed....

Thank you again!