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Python 如果数字高于上一个数字,则打印该数字。否则,请不要打印该数字_Python_Python 3.x - Fatal编程技术网

Python 如果数字高于上一个数字,则打印该数字。否则,请不要打印该数字

Python 如果数字高于上一个数字,则打印该数字。否则,请不要打印该数字,python,python-3.x,Python,Python 3.x,所以如果 它需要输出 mylist = [2,3,4,1,9,4] 因为3比2高,4比3高,1不比4高,这样继续下去……简单的解决方案怎么样?循环遍历数组并检查项目 [3,4,9] i=0 prevValue=0; 而我prevValue: 打印(prevValue) prevValue=mylist[i] 这是一种方法。使用简单的迭代和枚举 演示: i = 0 prevValue=0; while i < len(mylist ): if mylist[i] > pre

所以如果

它需要输出

mylist = [2,3,4,1,9,4]

因为3比2高,4比3高,1不比4高,这样继续下去……

简单的解决方案怎么样?循环遍历数组并检查项目

[3,4,9]
i=0
prevValue=0;
而我prevValue:
打印(prevValue)
prevValue=mylist[i]

这是一种方法。使用简单的迭代和
枚举

演示:

i = 0
prevValue=0;
while i < len(mylist ):
    if mylist[i] > prevValue:
        print(prevValue)
    prevValue=mylist[i]
mylist = [2,3,4,1,9,4]
res = []
for i, v in enumerate(mylist):
    try:
        if mylist[i+1] > v:
            res.append(mylist[i+1])
    except IndexError:
        pass
print(res)
输出:

i = 0
prevValue=0;
while i < len(mylist ):
    if mylist[i] > prevValue:
        print(prevValue)
    prevValue=mylist[i]
mylist = [2,3,4,1,9,4]
res = []
for i, v in enumerate(mylist):
    try:
        if mylist[i+1] > v:
            res.append(mylist[i+1])
    except IndexError:
        pass
print(res)

您可以使用类似于

这是一条单行线:

>>> it = iter(mylist)
>>> next(it, None)
2
>>> [y for x, y in zip(mylist, it) if y > x]
[3, 4, 9]
输出:

[i for idx, i in enumerate(mylist) if idx != 0 and i > mylist[idx - 1]]

所以我读了所有的答案,谢谢你抽出时间回答我的问题,我真的很感激。但是我找到了最适合我的答案,代码是:

[3, 4, 9]

到目前为止你试过什么?