Ejemplo n.º 1
0
def computeDiversityCoefficient(numberNodesInTree,sampleNameList,dataArray):
    #@sample is a list of (name of node,rank of node,sampleHitList)
    sample,_,numberNodes = takeNodesInTree(dataArray[7],sampleNameList)
    resultNodes = []
    for node in sample:
        assignments = 0
        if not (len(node) == 3):
            print " \n/!\ ERROR: List of samples wrong:",len(node),"."
            raise ValueError
        name,rank,sampleList = node[0],node[1],node[2]
        for hit in sampleList:
            if not (len(hit) == 2):
                print "\n/!\ ERROR: Sample Hit List wrong:",len(hit),"."
                raise ValueError
            assignments += hit[1]
        resultNodes.append(((name,rank),assignments))
    if not numberNodes:
        if not (len(sample)):
            print "\n/!\ ERROR: [BUG] [percentage/computeSamplesInAllMetadatum] Empty set of samples."
            raise ValueError
        return 0,resultNodes
    if not numberNodesInTree:
        print "\n/!\ ERROR: Taxonomic Tree is empty: whole tree:",numberNodesInTree,"."
        raise ValueError
    return (numberNodes/numberNodesInTree),resultNodes
Ejemplo n.º 2
0
def compute(tree,sampleNameList1,sampleNameList2):
    nodeList1,numberAT1,numberN1 = takeNodesInTree(tree,sampleNameList1)
    nodeList2,numberAT2,numberN2 = takeNodesInTree(tree,sampleNameList2)
    common, in1, in2 = [],[],[]
    #@numberC is the number of nodes in common
    numberC = 0
    for node in nodeList1:
        boolean,sampleHitList = memAndSampleHitList(node,nodeList2)
        #If @sampleHitList is not empty
        #That is if node also belongs to the second list of samples
        if boolean:
            common += [(node[0],node[1],node[2],sampleHitList)]
            numberC += 1
        else:
            in1 += [node]
    for node in nodeList1:
        #@boolean answers: Is it a node that does not belong to the first list of samples?
        boolean = True
        for cNode in common:
            if (node[0] == cNode[0] and node[1] == cNode[1]):
                boolean = False
        if boolean:
            in2 += [node]
    return common,in1,in2,numberAT1,numberAT2,numberN1,numberN2,numberC
Ejemplo n.º 3
0
def enumerateCommonPatterns(tree,sampleNameList1,sampleNameList2):
    commonPatternsList = []
    #@nodesList1 is the list of nodes (name,rank,sampleHitList) for sampleNameList1 (see misc.py)
    nodesList1,_,_ = takeNodesInTree(tree,sampleNameList1)
    #We consider every node of the tree as a potential root for a pattern
    for node in nodesList1:
        pattern = []
        numberAssignments = 0
        numberNodes = 0
        name,rank = node[0],node[1]
        #Gets the subtree (of the whole taxonomic tree) rooted at node
        root = tree.search(name,rank)
        #@candidateNodes is the list of TaxoTree nodes that can potentially be added to the pattern
        candidateNodes = [root]
        while candidateNodes:
            child = candidateNodes.pop()
            sampleHitList = child.sampleHitList
            #Checking if child has been assigned in both samples
            isInSampleList1 = []
            isInSampleList2 = []
            for x in sampleHitList:
                if inSample(x,sampleNameList1):
                    isInSampleList1.append(x)
                if inSample(x,sampleNameList2):
                    isInSampleList2.append(x)
            #If both lists are not empty, then node has been assigned in both samples
            if isInSampleList1 and isInSampleList2:
                #Merge the elements of both lists, deleting duplicates
                #e.g. if OPNA-J90 belongs to both sampleNameLists it would corrupt the result
                #as the assignments in this sample to child would be duplicated
                #(assuming there is no duplicate in each list)
                isInSample = mergeList(isInSampleList1,isInSampleList2)
                pattern.append((child.name,child.rank))
                numberNodes += 1
                for x in isInSample:
                    numberAssignments += x[1]
                candidateNodes += child.children
        #if the pattern is non-empty
        if pattern:
            commonPatternsList.append((pattern,numberAssignments,numberNodes))
    return commonPatternsList
Ejemplo n.º 4
0
def enumerateSpecificPatterns(tree,sampleNameListPattern,sampleNameListOther):
    specificPatternsList = []
    #List from where samples in both lists are deleted and only elements from sampleNameListPattern remain
    sampleNameListPatternTrimmed = trimList(sampleNameListPattern,sampleNameListOther)
    #@nodesList is the list of nodes (name,rank,sampleHitList) for sampleNameListPatternTrimmed
    nodesList,_,_ = takeNodesInTree(tree,sampleNameListPatternTrimmed)
    #Pretty much the same procedure than for @enumerateCommonPatterns
    for node in nodesList:
        pattern = []
        numberAssignments = 0
        numberNodes = 0
        name,rank = node[0],node[1]
        root = tree.search(name,rank)
        candidateNodes = [root]
        while candidateNodes:
            child = candidateNodes.pop()
            sampleHitList = child.sampleHitList
            isInSampleListPattern = []
            for x in sampleHitList:
                if inSample(x,sampleNameListPatternTrimmed):
                    isInSampleListPattern.append(x)
                if inSample(x,sampleNameListOther):
                    #This node is assigned in the samples of sampleNameListOther
                    #so it is discarded from the pattern
                    isInSamplePattern = []
                    break
            if isInSampleListPattern:
                pattern.append((child.name,child.rank))
                numberNodes += 1
                for x in isInSampleListPattern:
                    numberAssignments += x[1]
                candidateNodes += child.children
        #if the pattern is non-empty
        if pattern:
            specificPatternsList.append((pattern,numberAssignments,numberNodes))
    return specificPatternsList