将包含字符串和整数的列表中的元素求和,并将答案放入另一个列表python中

这是解决办法,你不必担心学生的分数。

list1=['student1',10,20,40,'student2',20,20,40,'student3',20,30,40,'student4',20,10,30]

student_name =[]
student_name_index =[]

for i in range(len(list1)):        
    if type(list1[i]) == int:
        pass
    else:
        student_name.append(list1[i])
        student_name_index.append(i)

student_name_index.append(len(list1)-1)

total_marks=[]

for i in range(1,len(student_name_index)):
    total_marks.append(sum(list1[student_name_index[i-1]+1:student_name_index[i]]))

final_result =[]

for name, mark in zip(student_name,total_marks):
    final_result.append(name)
    final_result.append(mark)

print(final_result)

# output ['student1', 70, 'student2', 80, 'student3', 90, 'student4', 30] 

尝试

in_lst = ['student1', 10, 20, 40, 'student2', 20, 20, 40, 'student3', 20, 30, 40, 'student4', 20, 10, 30]
out_lst = []
for x in range(0, len(in_lst), 4):
    student_entry = in_lst[x:x + 4]
    out_lst.append(student_entry[0])
    out_lst.append(sum(student_entry[1:]))
print(out_lst)

输出

['student1', 70, 'student2', 80, 'student3', 90, 'student4', 60]

它是这样的:0,4,8,所以不需要第二个for循环。

你已经知道号码在哪里了。(i+1,i+2,i+3)我是学生的名字。

list1=['student1',10,20,40,'student2',20,20,40,'student3',20,30,40,'student4',20,10,30]
list2=[]
for i in range(0, len(list1), 4):
    list2.append(list1[i])
    sum = list1[i+1]+list1[i+2]+list1[i+3]
    list2.append(sum)
print(list2)

list1=['student1',10,20,40,'student2',20,20,40,'student3',20,30,40,'student4',20,10,30]
list2=[]
for i in range(0,len(list1),4):
    list2.append(list1[i])
    sum= list1[i+1]+list1[i+2]+list1[i+3]
    list2.append(sum)
print(list2)

生产

['student1', 70, 'student2', 80, 'student3', 90, 'student4', 60]

代码中的第二个for循环遍历列表中的每个元素,并在索引mod 4为1时执行操作。你不需要两个for循环。

在第二个循环中,再次循环整个初始列表1,因此总是附加所有和。如果你的列表总是包含一个名字和三个等级,那么你可以去掉第二个循环,用第一个循环中的“i”作为迭代器来求和,就像你现在做的那样。

您可以使用单循环:

lst = ['student1', 10, 20, 40, 'student2', 20, 20, 40, 'student3', 20, 30, 40, 'student4', 20, 10, 30]
result = []
for i in range(0, len(lst), 4):
    result.extend((lst[i], sum(lst[i+1:i+4])))

输出:

['student1', 70, 'student2', 80, 'student3', 90, 'student4', 60]

如果每个学生的分数不同,例如:

lst = ['student1', 10, 20, 'student2', 10, 20, 30, 'student3', 10, 20, 30, 40, 'student4', 10, 20, 30, 40, 50]

然后:

s = 0
result = [lst[0]]
for i in lst[1:]:
    try:
        s += int(i)
    except ValueError:
        result.extend((s, i))
        s = 0
result.append(s)

输出:

['student1', 30, 'student2', 60, 'student3', 100, 'student4', 150]