Question

这是我必须转换为python的伪代码

A=99
LENGTH= LENGTH(list)
LIST= 92 50 26 82 73 
for P in range 0 to LENGTH-1
    IF LIST[P] <A THEN
        A=LIST[P]
        B=P
    ENDIF
ENDFOR
IF B < LENGTH THEN
    for P in range B to LENGTH -2
        LIST[P] = LIST[P+1]
    ENDFOR
ENDIF
LENGTH=LENGTH-1
LIST[LENGTH]=NULL

我在下面尝试进行编码，该代码旨在从LIST中删除最低值

a = 99
list=[92,50,26,82,73]

for  p in  range  (0,len(list) - 1):
    if list[p] < a :
        a = list[p] 
        b = p 

print (list) #I just added this to see what was happening

if  b < len(list):
    for p in range (b,len(list)-2):
        list[p]=list[p]+1

list=len(list)-1

print (list)
#I just added this to see what was happening

我已经编写了上面的代码，但并没有去除最小值

Answer 1

第一个列表是错误的变量名。它已经在python中具有了意义。
只需遍历列表项

for item in items:
    # use item.

如果您还想要索引

for idx in range(len(items)):   # no need for -1
    # use items[idx]

你的灵魂

minval = min(items)
new_items = []
for item in items:
    if item != minval:
        new_items.append(item)

# new items has the answer

Answer 2

您真的很亲密，以下是更正内容：

A = 99

l = [92, 50, 26, 82, 73]

LENGTH = len(l)

for P in range(LENGTH):
    if l[P] < A:
        A=l[P]
        B=P

if B < LENGTH:
    for P in range (B, LENGTH - 1):
        l[P] = l[P+1]


LENGTH=LENGTH-1
l[LENGTH]=None

现在尝试：

print(l) # [92, 50, 82, 73, None]

注意：由于Python使用基于0的索引，因此LENGTH-1和LENGTH-2分别更改为LENGTH和LENGTH-1。

Python删除最低值

2 个答案: