阻止值在每次字典迭代中递增

Question

我正在使用Python 2.7。我有两个tsv数据文件，我读入了两个字典，我想计算它们的recall分数，所以我需要计算tp和fn。 这些是我的字典的样子：

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

我的代码主要迭代gold字典，并删除gold字典末尾的数字key以及resultskey。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，我的代码似乎总是递增fn。这是我的可运行代码：

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：

tp = 1.00   fn =  3.00    recall = 0.25 

fn不正确，则应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

Answer 1

在我看来，您只想在结果中找不到匹配项时才增加fn。您可以使用变量来跟踪是否已找到匹配项，并在此基础上递增fn。在下面，我调整了您的代码并为此目的使用了match_found。

#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

Answer 2

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。

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('n-------------------------------------')
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