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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def get_probs_for_labels(labels, prediction_results):
 """ Given ML Workbench prediction results, get probs of each label for each instance. The prediction results are like: [ {'predicted': 'daisy', 'probability': 0.8, 'predicted_2': 'rose', 'probability_2': 0.1}, {'predicted': 'sunflower', 'probability': 0.9, 'predicted_2': 'daisy', 'probability_2': 0.01}, ] Each instance is ordered by prob. But in some cases probs are needed for fixed order of labels. For example, given labels = ['daisy', 'rose', 'sunflower'], the results of above is expected to be: [ [0.8, 0.1, 0.0], [0.01, 0.0, 0.9], ] Note that the sum of each instance may not be always 1. If model's top_n is set to none-zero, and is less than number of labels, then prediction results may not contain probs for all labels. Args: labels: a list of labels specifying the order of the labels. prediction_results: a pandas DataFrame containing prediction results, usually returned by get_prediction_results() call. Returns: A list of list of probs for each class. """


  probs = []
  if 'probability' in prediction_results:
    # 'probability' exists so top-n is set to none zero, and results are like
    # "predicted, predicted_2,...,probability,probability_2,...
    for i, r in prediction_results.iterrows():
      probs_one = [0.0] * len(labels)
      for k, v in six.iteritems(r):
        if v in labels and k.startswith('predicted'):
          if k == 'predict':
            prob_name = 'probability'
          else:
            prob_name = 'probability' + k[9:]
          probs_one[labels.index(v)] = r[prob_name]
      probs.append(probs_one)
    return probs
  else:
    # 'probability' does not exist, so top-n is set to zero. Results are like
    # "predicted, class_name1, class_name2,...
    for i, r in prediction_results.iterrows():
      probs_one = [0.0] * len(labels)
      for k, v in six.iteritems(r):
        if k in labels:
          probs_one[labels.index(k)] = v
      probs.append(probs_one)
    return probs