예제 #1
0
def exp11():
    """
    Show correlation for each feature
    """
    poems = getPoemModel().poems
    scores = {}
    scores["affect"] = getAffectRatios()
    scores["cLength"] = getLogAverageCommentLength()
    scores["rating"] = getPoemScores()
    scores["typeToken"] = getCommentTypeTokenRatio(100)
    scores["numC"] = getNumberOfComments(True)  # use log

    result = {}
    for k1, v1 in scores.items():
        for feature in poems.values()[0].keys():
            cor, p = getCorrelation(poems, v1, feature)
            if result.get(feature, None) is None:
                result[feature] = {k1: (cor, p)}
            else:
                result[feature][k1] = (cor, p)

    for k1 in sorted(result.keys(), key=lambda x: result[x]["affect"][1]):  # sort by affect
        print "\\\\", k1, "& %0.2f & %0.4f" % result[k1]["affect"], "& %0.2f & %0.4f" % result[k1][
            "typeToken"
        ], "& %0.2f & %0.4f" % result[k1]["cLength"], "& %0.2f & %0.4f" % result[k1][
            "rating"
        ], "& %0.2f & %0.4f" % result[
            k1
        ][
            "numC"
        ]
예제 #2
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def exp06():
    """
    Can we predict the log of the number of responses? Not much better.
    """
    m = getPoemModel()
    poems = m.poems
    scores = getNumberOfComments(True)  # use log
    makePlots(poems, scores, "log of number of comments", "../experiments/exp06.pdf")
    runPredictCV(poems, scores, DEFAULT_FEATURE_LIST)
예제 #3
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def exp05():
    """
    Can we predict the number of responses? Not so well.
    """
    m = getPoemModel()
    poems = m.poems
    scores = getNumberOfComments()
    makePlots(poems, scores, "number of comments", "../experiments/exp05.pdf")
    runPredictCV(poems, scores, DEFAULT_FEATURE_LIST)