From 4759c20103e20ff2737e9fcd8b3fbe660717469f Mon Sep 17 00:00:00 2001
From: "weiye.wang" <kj_wangweiye@126.com>
Date: Tue, 18 Jun 2024 06:55:19 +0800
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---
 工具v4/database_tools_2.py | 25 ++++++++++++++-----------
 1 file changed, 14 insertions(+), 11 deletions(-)

diff --git a/工具v4/database_tools_2.py b/工具v4/database_tools_2.py
index de9e4a12..3d0dcdf8 100644
--- a/工具v4/database_tools_2.py
+++ b/工具v4/database_tools_2.py
@@ -2732,19 +2732,22 @@ def CheckPaperType(filepath,filename): #根据filepath(通常是小闲的zip解
 def generateColIndexandMarks(filepath,statsfilename,paperinfo): #根据filepath(是一个有statsfilename的文件夹列表)中第一个路径中的数据文件, statsfilename数据文件名, 及paperinfo(FindPaper返回的结果)寻找excel文件中有效的列的位置和相应的满分分数
     dir = filepath[0]
     dfcurrent = pd.read_excel(os.path.join(dir,statsfilename))
-    validcols = []
+    excludecols = []
     for i in range(len(dfcurrent.columns)):
         colname = str(dfcurrent.iloc[1,i])
-        if ("单选" in colname or "多选" in colname or "填空" in colname or "主观" in colname or "步" in colname ) and re.findall("[ABCD]",colname) == []:
-            validcols.append(i)
-    for col in range(len(validcols)-1,-1,-1):
-        colname = str(dfcurrent.iloc[1,validcols[col]])
-        if "主观" in colname:
-            colname_main = re.findall(r"^([\d\.]*)[\($]",colname[2:])[0]
-            t = [dfcurrent.iloc[1,c] for c in validcols[col+1:]]
-            t = str(t)
-            if colname_main in t:
-                validcols.pop(col)
+        if re.findall("[A-Z]",colname) != []:
+            excludecols.append(i)
+    validcoldict = {}
+    for i in range(len(dfcurrent.columns)):
+        colname = str(dfcurrent.iloc[1,i])
+        digname = re.findall(r"[\d\.]+",colname)
+        if len(digname) > 0 and not i in excludecols:
+            if not i-1 in validcoldict or not validcoldict[i-1] in digname[0]:
+                validcoldict[i] = digname[0]
+            else:
+                validcoldict.pop(i-1)
+                validcoldict[i] = digname[0]
+    validcols = list(validcoldict.keys())
     if paperinfo[3] == []:
         marks = [1] * len(validcols)
     else: