关于python:阻止一个值在每次字典迭代中递增

Stop a value from being incremented in each dictionary iteration

我使用的是 Python 2.7。我有两个 tsv 数据文件,我将它们读入两个字典,我想计算它们的 recall 分数,所以我需要计算 tpfn
这些是我的字典的样子:

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gold = {'A11':'cat', 'A22':'cat', 'B3':'mouse'}
results = {'A2':'cat', 'B2':'dog'}

我的代码主要是迭代gold字典,去掉gold字典keyresultskey末尾的数字。然后,检查键是否匹配以查找它们的值是否匹配以计算 tp。但是,我的代码似乎总是增加 fn。这是我的可运行代码:

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from __future__ import division
import string


def eval():
        tp=0 #true positives
        fn=0 #false negatives
        fp=0#false positives

        gold = {'A11':'cat', 'A22':'cat', 'B3':'mouse'}
        results = {'A2':'cat', 'B2':'dog'}

       #iterate gold dictionary
        for i,j in gold.items():

            #remove the digits off gold keys
            i_stripped = i.rstrip(string.digits)

            #iterate results dictionary
            for k,v in results.items():

                #remove the digits off results keys
                k_stripped = k.rstrip(string.digits)

                # check if key match!
                if i_stripped == k_stripped:

                  #check if values match then increment tp
                  if j == v:
                      tp += 1

                      #delete dictionary entries to avoid counting them again
                      del gold_copy[i]
                      del results_copy[k]

                      #get out of this loop we found a match!
                      break
                continue

            # NO match was found in the results, then consider it as fn
            fn += 1 #<------ wrong calculations caused in this line

        print 'tp = %.2f   fn =  %.2f    recall = %.2f ' % (tp, fn, float(tp)/(tp+fn))

这是输出:

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tp = 1.00   fn =  3.00    recall = 0.25

fn 不正确,应该是 2 而不是 3。如何阻止 fn 在每次迭代中递增?任何指导将不胜感激。

谢谢你,


在我看来,只有在结果中没有找到匹配项时,您才想增加 fn。您可以使用变量来跟踪是否已找到匹配项,并在此基础上增加 fn。在下面,我调整了您的代码并为此目的使用了 match_found

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#iterate gold dictionary
 for i,j in gold.items():

     # create a variable that indicates whether a match was found
     match_found = False

     #remove the digits off gold keys
     i_stripped = i.rstrip(string.digits)

     #iterate results dictionary
     for k,v in results.items():

         #remove the digits off results keys
         k_stripped = k.rstrip(string.digits)

         # check if key match!
         if i_stripped == k_stripped:

           #check if values match then increment tp
           if j == v:
               tp += 1

               # now a match has been found, change variable
               match_found = True

               #delete dictionary entries to avoid counting them again
               del gold_copy[i]
               del results_copy[k]

               #get out of this loop we found a match!
               break
         continue

     # NO match was found in the results, then consider it as fn
     # now, only if no match has been found, increment fn
     if not match_found :
         fn += 1 #<------ wrong calculations caused in this line


如果这不是您所需要的,您应该能够对其进行修改以使其正常工作。

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tp = 0 #true positives
fn = 0 #false negatives
fp = 0 #false positives


gold = {'A11':'cat', 'A22':'cat', 'B3':'mouse'}
results = {'A2':'cat', 'B2':'dog'}

for gold_k, gold_v in gold.items():
    # Remove digits and make lower case
    clean_gold_k = gold_k.rstrip(string.digits).lower()

    for results_k, results_v in results.items():
        # Remove digits and make lower case
        clean_results_k = results_k.rstrip(string.digits).lower()

        keys_agree = clean_gold_k == clean_results_k
        values_agree = gold_v.lower() == results_v.lower()

        print('\
-------------------------------------'
)
        print('Gold = ' + gold_k + ': ' + gold_v)
        print('Result = ' + results_k + ': ' + results_v)

        if keys_agree and values_agree:
            print('tp')
            tp += 1

        elif keys_agree and not values_agree:
            print('fn')
            fn += 1

        elif values_agree and not keys_agree:
            print('fp')
            fp += 1