阻止值在每次字典迭代中递增
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recall它们的分数，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'}
我的代码主要是迭代的gold字典，并在年底消除数字gold字典key以及results key。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：
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是不正确的，应该是2而不是3。如何停止fn在每次迭代中递增？任何指导将不胜感激。
谢谢，
ITMISS2回答
-
森栏
在我看来，您只想fn在结果中找不到匹配项时才增加。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用于match_found此目的。#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 -
HUX布斯
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 positivesfn = 0 #false negativesfp = 0 #false positivesgold = {'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
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