if itme1 == '总计': count = count_conditions( teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '女': count = count_conditions( teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '少数民族': count = count_conditions( teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '在编人员': count = count_conditions( teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count优化代码

可以将重复的代码段封装成一个函数，传入不同的参数来实现不同的功能，避免重复代码。例如：

def count_conditions(teacher_arr, condition_func, time_func, count):
    count = count_conditions(teacher_arr, condition_func, lambda row: row[8] == itme)
    if count != 0:
        sh_4149.cell(index1 6, index 4).value = count

# 使用方法
count_conditions(teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[2] == '女', '女')
count_conditions(teacher_arr, lambda row: row[11] != '汉族', lambda row: row[12] == '是', '少数民族')

相关问题

for index1, itme1 in enumerate(list_col): for index, itme in enumerate(list_row): if itme1 == '总计': count = count_conditions( teacher_arr, lambda row: row[9] == '学前' or row[9] == '小学', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '女': count = count_conditions( teacher_arr, lambda row: row[2] == '女', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '少数民族': count = count_conditions( teacher_arr, lambda row: row[11] != '汉族', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count if itme1 == '在编人员': count = count_conditions( teacher_arr, lambda row: row[12] == '是', lambda row: row[8] == itme) if count != 0: sh_4149.cell(index1+6, index+4).value = count代码优化

：
首先，可以将每一个if语句的判断条件都提前判断一遍，如果不满足就直接跳过，避免重复计算和赋值。
其次，可以将count_conditions函数的每个参数使用lambda函数封装成一个函数参数，以简化代码。
最后，可以将重复的赋值操作提取出来，避免重复代码。
优化后的代码如下：

count_conditions_filter1 = lambda row: row[9] == '学前' or row[9] == '小学'
count_conditions_filter2 = lambda row: row[2] == '女'
count_conditions_filter3 = lambda row: row[11] != '汉族'
count_conditions_filter4 = lambda row: row[12] == '是'

for index1, item1 in enumerate(list_col):
count = 0
if item1 == '总计':
count = count_conditions(teacher_arr, count_conditions_filter1, lambda row: row[8] == item)
elif item1 == '女':
count = count_conditions(teacher_arr, count_conditions_filter2, lambda row: row[8] == item)
elif item1 == '少数民族':
count = count_conditions(teacher_arr, count_conditions_filter3, lambda row: row[8] == item)
elif item1 == '在编人员':
count = count_conditions(teacher_arr, count_conditions_filter4, lambda row: row[8] == item)

if count != 0:
sh_4149.cell(index1 6, index 4).value = count

if itme1 == '小学': count = count_conditions( change_student_arr, lambda row: row[6] == itme1, lambda row: row[4] == itme) if count != 0: sh_3040.cell(index1+7, index+6).value = count if itme1 == '女': count = count_conditions( change_student_arr, lambda row: row[2] == itme1, lambda row: row[4] == itme) if count != 0: sh_3040.cell(index1+7, index+6).value = count if itme1 == '少数民族': count = count_conditions( change_student_arr, lambda row: row[3] == itme1, lambda row: row[4] == itme) if count != 0: sh_3040.cell(index1+7, index+6).value = count代码优化

ChitGPT：您好！关于您的代码，我可以提供一些改进意见。首先，可以将多个if语句合并成一个，以减少冗余代码。具体来说，请尝试以下改进：

count = 0
if itme1 == '小学':
count = count_conditions(change_student_arr, lambda row: row[6] == itme1, lambda row: row[4] == itme)
elif itme1 == '女':
count = count_conditions(change_student_arr, lambda row: row[2] == itme1, lambda row: row[4] == itme)
elif itme1 == '少数民族':
count = count_conditions(change_student_arr, lambda row: row[3] == itme1, lambda row: row[4] == itme)

if count != 0:
sh_3040.cell(index1 7, index 6).value = count

这样可以让代码更加简洁清晰，同时还能提高代码的可读性和可维护性。如果您有任何问题或者需要我帮忙进行其他的改进，欢迎随时告诉我。