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Concatenating dictionary values and keys, and further operations

P: n/a
I wrote the following code to concatenate every 2 keys of a dictionary and
their corresponding values.
e.g if i have tiDict1 = tiDict1 = {'a':[1,2],'b':[3,4,5]} i should get
tiDict2={'ab':[1,2][3,4,5]} and similarly for dicts with larger no. of
features.
Now i want to check each pair to see if they are connected...element of
this pair will be one from the first list and one from the second....e.g
for 'ab' i want to check if 1 and 3 are connected,then 1 and 4,then 1 and
5,then 2 and 3,then 2 and 4,then 2 and 5.
The information of this connected thing is in a text file as follows:
1,'a',2,'b'
3,'a',5,'a'
3,'a',6,'a'
3,'a',7,'b'
8,'a',7,'b'
..
..
This means 1(type 'a') and 2(type 'b') are connected,3 and 5 are connected
and so on.
I am not able to figure out how to do this.Any pointers would be helpful
Here is the code i have written till now:
Expand|Select|Wrap|Line Numbers
  1. def genTI(tiDict):
  2. tiDict1 = {}
  3. tiList = [tiDict1.keys(),tiDict1.values()]
  4. length =len(tiDict1.keys())-1
  5. for i in range(0,length,1):
  6. for j in range(0,length,1):
  7. for k in range(1,length+1,1):
  8. if j+k <= length:
  9. key = tiList[i][j] + tiList[i][j+k]
  10. value = [tiList[i+1][j],tiList[i+1][j+k]]
  11. tiDict2[key] = value
  12. continue
  13. continue
  14. continue
  15. return tiDict2
  16.  
Thanks in advance,
girish
Jun 5 '06 #1
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11 Replies


P: n/a
Girish Sahani wrote:
I wrote the following code to concatenate every 2 keys of a dictionary and
their corresponding values.
e.g if i have tiDict1 = tiDict1 = {'a':[1,2],'b':[3,4,5]} i should get
tiDict2={'ab':[1,2][3,4,5]} and similarly for dicts with larger no. of
features.
Now i want to check each pair to see if they are connected...element of
this pair will be one from the first list and one from the second....e.g
for 'ab' i want to check if 1 and 3 are connected,then 1 and 4,then 1 and
5,then 2 and 3,then 2 and 4,then 2 and 5.
The information of this connected thing is in a text file as follows:
1,'a',2,'b'
3,'a',5,'a'
3,'a',6,'a'
3,'a',7,'b'
8,'a',7,'b'
.
.
This means 1(type 'a') and 2(type 'b') are connected,3 and 5 are connected
and so on.
I am not able to figure out how to do this.Any pointers would be helpful

Girish

It seems you want the Cartesian product of every pair of lists in the
dictionary, including the product of lists with themselves (but you
don't say why ;-)).

I'm not sure the following is exactly what you want or if it is very
efficient, but maybe it will start you off. It uses a function
'xcombine' taken from a recipe in the ASPN cookbook by David
Klaffenbach (2004).

(It should give every possibility, which you then check in your file)

Gerard

-------------------------------------------------------------------------

def nkRange(n,k):
m = n - k + 1
indexer = range(0, k)
vector = range(1, k+1)
last = range(m, n+1)
yield vector
while vector != last:
high_value = -1
high_index = -1
for i in indexer:
val = vector[i]
if val > high_value and val < m + i:
high_value = val
high_index = i
for j in range(k - high_index):
vector[j+high_index] = high_value + j + 1
yield vector

def kSubsets( alist, k ):
n = len(alist)
for vector in nkRange(n, k):
ret = []
for i in vector:
ret.append( alist[i-1] )
yield ret

data = { 'a': [1,2], 'b': [3,4,5], 'c': [1,4,7] }

pairs = list( kSubsets(data.keys(),2) ) + [ [k,k] for k in
data.iterkeys() ]
print pairs
for s in pairs:
for t in xcombine( data[s[0]], data[s[1]] ):
print "%s,'%s',%s,'%s'" % ( t[0], s[0], t[1], s[1] )
-------------------------------------------------------------------------

1,'a',1,'c'
1,'a',4,'c'
1,'a',7,'c'
2,'a',1,'c'
2,'a',4,'c'
2,'a',7,'c'
1,'a',3,'b'
1,'a',4,'b'
1,'a',5,'b'
2,'a',3,'b'
2,'a',4,'b'
2,'a',5,'b'
1,'c',3,'b'
1,'c',4,'b'
1,'c',5,'b'
4,'c',3,'b'
4,'c',4,'b'
4,'c',5,'b'
7,'c',3,'b'
7,'c',4,'b'
7,'c',5,'b'
1,'a',1,'a'
1,'a',2,'a'
2,'a',1,'a'
2,'a',2,'a'
1,'c',1,'c'
1,'c',4,'c'
1,'c',7,'c'
4,'c',1,'c'
4,'c',4,'c'
4,'c',7,'c'
7,'c',1,'c'
7,'c',4,'c'
7,'c',7,'c'
3,'b',3,'b'
3,'b',4,'b'
3,'b',5,'b'
4,'b',3,'b'
4,'b',4,'b'
4,'b',5,'b'
5,'b',3,'b'
5,'b',4,'b'
5,'b',5,'b'

Jun 5 '06 #2

P: n/a

Gerard Flanagan wrote:
Girish Sahani wrote:
I wrote the following code to concatenate every 2 keys of a dictionary and
their corresponding values.
e.g if i have tiDict1 = tiDict1 = {'a':[1,2],'b':[3,4,5]} i should get
tiDict2={'ab':[1,2][3,4,5]} and similarly for dicts with larger no. of
features.
Now i want to check each pair to see if they are connected...element of
this pair will be one from the first list and one from the second....e.g
for 'ab' i want to check if 1 and 3 are connected,then 1 and 4,then 1 and
5,then 2 and 3,then 2 and 4,then 2 and 5.
The information of this connected thing is in a text file as follows:
1,'a',2,'b'
3,'a',5,'a'
3,'a',6,'a'
3,'a',7,'b'
8,'a',7,'b'
.
.
This means 1(type 'a') and 2(type 'b') are connected,3 and 5 are connected
and so on.
I am not able to figure out how to do this.Any pointers would be helpful

Girish

It seems you want the Cartesian product of every pair of lists in the
dictionary, including the product of lists with themselves (but you
don't say why ;-)).

I'm not sure the following is exactly what you want or if it is very
efficient, but maybe it will start you off. It uses a function
'xcombine' taken from a recipe in the ASPN cookbook by David
Klaffenbach (2004).


http://aspn.activestate.com/ASPN/Coo.../Recipe/302478

Jun 5 '06 #3

P: n/a
I have a text file in the following format:

1,'a',2,'b'
3,'a',5,'c'
3,'a',6,'c'
3,'a',7,'b'
8,'a',7,'b'
..
..
..
Now i need to generate 2 things by reading the file:
1) A dictionary with the numbers as keys and the letters as values.
e.g the above would give me a dictionary like
{1:'a', 2:'b', 3:'a', 5:'c', 6:'c' ........}
2) A list containing pairs of numbers from each line.
The above formmat would give me the list as
[[1,2],[3,5],[3,6][3,7][8,7]......]

I wrote the following codes for both of these but the problem is that
lines returns a list like ["1,'a',2,'b'","3,'a',5,'c","3,'a',6,'c'".....]
Now due to the "" around each line,it is treated like one object
and i cannot access the elements of a line.

Expand|Select|Wrap|Line Numbers
  1. #code to generate the dictionary
  2. def get_colocations(filename):
  3. lines = open(filename).read().split("\n")
  4. colocnDict = {}
  5. i = 0
  6. for line in lines:
  7. if i <= 2:
  8. colocnDict[line[i]] = line[i+1]
  9. i+=2
  10. continue
  11. return colocnDict
  12.  
Expand|Select|Wrap|Line Numbers
  1. def genPairs(filename):
  2. lines = open(filename).read().split("\n")
  3. pairList = []
  4. for line in lines:
  5. pair = [line[0],line[2]]
  6. pairList.append(pair)
  7. i+=2
  8. continue
  9. return pairList
  10.  
Please help :((
Jun 6 '06 #4

P: n/a
Girish Sahani wrote:
1) A dictionary with the numbers as keys and the letters as values.
e.g the above would give me a dictionary like
{1:'a', 2:'b', 3:'a', 5:'c', 6:'c' ........}
def get_dict( f ) :
out = {}
for line in file(f) :
n1,s1,n2,s2 = line.split(',')
out.update( { int(n1):s1[1], int(n2):s2[1] } )
return out
2) A list containing pairs of numbers from each line.
The above formmat would give me the list as
[[1,2],[3,5],[3,6][3,7][8,7]......]


def get_pairs( f ) :
out = []
for line in file(f) :
n1,_,n2,_ = line.split(',')
out.append( [int(n1),int(n2)] )
return out

Regards
Sreeram
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Jun 6 '06 #5

P: n/a
On 6/06/2006 2:10 PM, Girish Sahani wrote:
I have a text file in the following format:

1,'a',2,'b'
3,'a',5,'c'
3,'a',6,'c'
3,'a',7,'b'
8,'a',7,'b'
Check out the csv module.
.
.
.
Now i need to generate 2 things by reading the file:
1) A dictionary with the numbers as keys and the letters as values.
e.g the above would give me a dictionary like
{1:'a', 2:'b', 3:'a', 5:'c', 6:'c' ........}
2) A list containing pairs of numbers from each line.
The above formmat would give me the list as
[[1,2],[3,5],[3,6][3,7][8,7]......]

I wrote the following codes for both of these but the problem is that
lines returns a list like ["1,'a',2,'b'","3,'a',5,'c","3,'a',6,'c'".....]
Now due to the "" around each line,it is treated like one object
and i cannot access the elements of a line.
You managed to split the file contents into lines using
lines = open(filename).read().split("\n")
Same principle applies to each line:

|>>> lines = ["1,'a',2,'b'","3,'a',5,'c","3,'a',6,'c'"]
|>>> lines[0].split(',')
['1', "'a'", '2', "'b'"]
|>>> lines[1].split(',')
['3', "'a'", '5', "'c"]
|>>>

Expand|Select|Wrap|Line Numbers
  1.  #code to generate the dictionary
  2.  def get_colocations(filename):
  3.      lines = open(filename).read().split("\n")
  4.      colocnDict = {}
  5.      i = 0
  6.      for line in lines:
  7.          if i <= 2:
  8.              colocnDict[line[i]] = line[i+1]
  9.              i+=2
  10.              continue
  11.          return colocnDict
  • The return is indented too far; would return after 1st line.
  •  

  • Expand|Select|Wrap|Line Numbers
    1.  def genPairs(filename):
    2.      lines = open(filename).read().split("\n")
    3.      pairList = []
    4.      for line in lines:
    5.          pair = [line[0],line[2]]
    6.          pairList.append(pair)
    7.          i+=2
    8.  
    9. i is not defined. This would cause an exception. Please *always* post
    10. the code that you actually ran.
    11.          continue
    12.  return pairList
    13.  
    14. dedented too far!!
    15.  
    Please help :((

    def get_both(filename):
    lines = open(filename).read().split("\n")
    colocnDict = {}
    pairList = []
    for line in lines:
    n1, b1, n2, b2 = line.split(",")
    n1 = int(n1)
    n2 = int(n2)
    a1 = b1.strip("'")
    a2 = b2.strip("'")
    colocnDict[n1] = a1
    colocnDict[n2] = a2
    pairList.append([n1, n2])
    return colocnDict, pairList

    def get_both_csv(filename):
    import csv
    reader = csv.reader(open(filename, "rb"), quotechar="'")
    colocnDict = {}
    pairList = []
    for n1, a1, n2, a2 in reader:
    n1 = int(n1)
    n2 = int(n2)
    colocnDict[n1] = a1
    colocnDict[n2] = a2
    pairList.append([n1, n2])
    return colocnDict, pairList

    HTH,
    John
    Jun 6 '06 #6

    P: n/a
    Girish Sahani wrote:
    Gerard Flanagan wrote:
    Girish Sahani wrote:
    > I wrote the following code to concatenate every 2 keys of a dictionary and
    > their corresponding values.
    > e.g if i have tiDict1 = tiDict1 = {'a':[1,2],'b':[3,4,5]} i should get
    > tiDict2={'ab':[1,2][3,4,5]} and similarly for dicts with larger no. of
    > features.
    > Now i want to check each pair to see if they are connected...element

    of
    > this pair will be one from the first list and one from the

    second....e.g
    > for 'ab' i want to check if 1 and 3 are connected,then 1 and 4,then 1

    and
    > 5,then 2 and 3,then 2 and 4,then 2 and 5.
    > The information of this connected thing is in a text file as follows:
    > 1,'a',2,'b'
    > 3,'a',5,'a'
    > 3,'a',6,'a'
    > 3,'a',7,'b'
    > 8,'a',7,'b'
    > .
    > .
    > This means 1(type 'a') and 2(type 'b') are connected,3 and 5 are

    connected
    > and so on.
    > I am not able to figure out how to do this.Any pointers would be

    helpful
    Girish

    It seems you want the Cartesian product of every pair of lists in the
    dictionary, including the product of lists with themselves (but you
    don't say why ;-)).

    I'm not sure the following is exactly what you want or if it is very
    efficient, but maybe it will start you off. It uses a function
    'xcombine' taken from a recipe in the ASPN cookbook by David
    Klaffenbach (2004).


    http://aspn.activestate.com/ASPN/Coo.../Recipe/302478

    --
    http://mail.python.org/mailman/listinfo/python-list

    Thanks a lot Gerard and Roberto.but i think i should explain the exact
    thing with an example.
    Roberto what i have right now is concatenating the keys and the
    corresponding values:
    e.g {'a':[1,2],'b':[3,4,5],'c':[6,7]} should give me
    {'ab':[1,2][3,4,5] 'ac':[1,2][6,7] 'bc':[3,4,5][6,7]}
    The order doesnt matter here.It could be 'ac' followed by 'bc' and 'ac'.
    Also order doesnt matter in a string:the pair 'ab':[1,2][3,4,5] is same as
    'ba':[3,4,5][1,2].
    This representation means 'a' corresponds to the list [1,2] and 'b'
    corresponds to the list [3,4,5].
    Now, for each key-value pair,e.g for 'ab' i must check each feature in the
    list of 'a' i.e. [1,2] with each feature in list of 'b' i.e. [3,4,5].So I
    want to take cartesian product of ONLY the 2 lists [1,2] and [3,4,5].
    Finally i want to check each pair if it is present in the file,whose
    format i had specified.
    The code Gerard has specified takes cartesian products of every 2 lists.
    Hi Garish,

    it's better to reply to the Group.
    Now, for each key-value pair,e.g for 'ab' i must check each feature in the
    list of 'a' i.e. [1,2] with each feature in list of 'b' i.e. [3,4,5].So I
    want to take cartesian product of ONLY the 2 lists [1,2] and [3,4,5].


    I'm confused. You say *for each* key-value pair, and you wrote above
    that the keys were the 'concatenation' of "every 2 keys of a
    dictionary".

    Sorry, too early for me. Maybe if you list every case you want, given
    the example data.

    All the best.

    Gerard

    Jun 6 '06 #7

    P: n/a
    Really sorry for that indentation thing :)
    I tried out the code you have given, and also the one sreeram had written.
    In all of these,i get the same error of this type:
    Error i get in Sreeram's code is:
    n1,_,n2,_ = line.split(',')
    ValueError: need more than 1 value to unpack

    And error i get in your code is:
    for n1, a1, n2, a2 in reader:
    ValueError: need more than 0 values to unpack

    Any ideas why this is happening?

    Thanks a lot,
    girish

    Jun 6 '06 #8

    P: n/a
    On 6/06/2006 4:15 PM, Girish Sahani wrote:
    Really sorry for that indentation thing :)
    I tried out the code you have given, and also the one sreeram had written.
    In all of these,i get the same error of this type:
    Error i get in Sreeram's code is:
    n1,_,n2,_ = line.split(',')
    ValueError: need more than 1 value to unpack

    And error i get in your code is:
    for n1, a1, n2, a2 in reader:
    ValueError: need more than 0 values to unpack

    Any ideas why this is happening?
    In the case of my code, this is consistent with the line being empty,
    probably the last line. As my mentor Bruno D. would say, your test data
    does not match your spec :-) Which do you want to change, the spec or
    the data?

    You can change my csv-reading code to detect dodgy data like this (for
    example):

    for row in reader:
    if not row:
    continue # ignore empty lines, wherever they appear
    if len(row) != 4:
    raise ValueError("Malformed row %r" % row)
    n1, a1, n2, a2 = row

    In the case of Sreeram's code, perhaps you could try inserting
    print "line = ", repr(line)
    before the statement that is causing the error.


    Thanks a lot,
    girish

    Jun 6 '06 #9

    P: n/a

    Girish> I have a text file in the following format:
    Girish> 1,'a',2,'b'
    Girish> 3,'a',5,'c'
    Girish> 3,'a',6,'c'
    Girish> 3,'a',7,'b'
    Girish> 8,'a',7,'b'
    Girish> .
    Girish> .
    Girish> .
    Girish> Now i need to generate 2 things by reading the file:
    Girish> 1) A dictionary with the numbers as keys and the letters as values.
    Girish> e.g the above would give me a dictionary like
    Girish> {1:'a', 2:'b', 3:'a', 5:'c', 6:'c' ........}
    Girish> 2) A list containing pairs of numbers from each line.
    Girish> The above formmat would give me the list as
    Girish> [[1,2],[3,5],[3,6][3,7][8,7]......]

    Running this:

    open("some.text.file", "w").write("""\
    1,'a',2,'b'
    3,'a',5,'c'
    3,'a',6,'c'
    3,'a',7,'b'
    8,'a',7,'b'
    """)

    import csv

    class dialect(csv.excel):
    quotechar = "'"
    reader = csv.reader(open("some.text.file", "rb"), dialect=dialect)
    mydict = {}
    mylist = []
    for row in reader:
    numbers = [int(n) for n in row[::2]]
    letters = row[1::2]
    mydict.update(dict(zip(numbers, letters)))
    mylist.append(numbers)

    print mydict
    print mylist

    import os

    os.unlink("some.text.file")

    displays this:

    {1: 'a', 2: 'b', 3: 'a', 5: 'c', 6: 'c', 7: 'b', 8: 'a'}
    [[1, 2], [3, 5], [3, 6], [3, 7], [8, 7]]

    That seems to be approximately what you're looking for.

    Skip
    Jun 6 '06 #10

    P: n/a
    > On 6/06/2006 4:15 PM, Girish Sahani wrote:
    Really sorry for that indentation thing :)
    I tried out the code you have given, and also the one sreeram had
    written.
    In all of these,i get the same error of this type:
    Error i get in Sreeram's code is:
    n1,_,n2,_ = line.split(',')
    ValueError: need more than 1 value to unpack

    And error i get in your code is:
    for n1, a1, n2, a2 in reader:
    ValueError: need more than 0 values to unpack

    Any ideas why this is happening?
    In the case of my code, this is consistent with the line being empty,
    probably the last line. As my mentor Bruno D. would say, your test data
    does not match your spec :-) Which do you want to change, the spec or
    the data?

    Thanks John, i just changed my Data file so as not to contain any empty
    lines, i guess that was the easier solution ;)
    You can change my csv-reading code to detect dodgy data like this (for
    example):

    for row in reader:
    if not row:
    continue # ignore empty lines, wherever they appear
    if len(row) != 4:
    raise ValueError("Malformed row %r" % row)
    n1, a1, n2, a2 = row

    In the case of Sreeram's code, perhaps you could try inserting
    print "line = ", repr(line)
    before the statement that is causing the error.


    Thanks a lot,
    girish

    --
    http://mail.python.org/mailman/listinfo/python-list


    Jun 7 '06 #11

    P: n/a
    Girish said, through Gerard's forwarded message:
    Thanks a lot Gerard and Roberto.but i think i should explain the exact
    thing with an example.
    Roberto what i have right now is concatenating the keys and the
    corresponding values:
    e.g {'a':[1,2],'b':[3,4,5],'c':[6,7]} should give me
    {'ab':[1,2][3,4,5] 'ac':[1,2][6,7] 'bc':[3,4,5][6,7]}
    The order doesnt matter here.It could be 'ac' followed by 'bc' and 'ac'.
    Also order doesnt matter in a string:the pair 'ab':[1,2][3,4,5] is same as
    'ba':[3,4,5][1,2].
    This representation means 'a' corresponds to the list [1,2] and 'b'
    corresponds to the list [3,4,5].
    The problem if that the two lists aren't distinguishable when
    concatenated, so what you get is [1, 2, 3, 4, 5]. You have to pack
    both lists in a tuple: {'ab': ([1, 2], [3, 4, 5]), ...}
    d = {'a':[1, 2], 'b':[3, 4, 5], 'c':[6, 7]}
    d2 = dict(((i + j), (d[i], d[j])) for i in d for j in d if i < j)
    d2 {'ac': ([1, 2], [6, 7]), 'ab': ([1, 2], [3, 4, 5]), 'bc': ([3, 4, 5], [6, 7])}
    Now, for each key-value pair,e.g for 'ab' i must check each feature in the
    list of 'a' i.e. [1,2] with each feature in list of 'b' i.e. [3,4,5].So I
    want to take cartesian product of ONLY the 2 lists [1,2] and [3,4,5].
    You can do this without creating an additional dictionary:
    d = {'a':[1, 2], 'b':[3, 4, 5], 'c':[6, 7]}
    pairs = [i + j for i in d for j in d if i < j]
    for i, j in pairs:

    .... cartesian_product = [(x, y) for x in d[i] for y in d[j]]
    .... print i + j, cartesian_product
    ....
    ac [(1, 6), (1, 7), (2, 6), (2, 7)]
    ab [(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5)]
    bc [(3, 6), (3, 7), (4, 6), (4, 7), (5, 6), (5, 7)]

    You can do whatever you want with this cartesian product inside the loop.
    Finally i want to check each pair if it is present in the file,whose
    format i had specified.


    I don't understand the semantics of the file format, so I leave this
    as an exercise to the reader :)
    Best regards.
    --
    Roberto Bonvallet
    Jun 7 '06 #12

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