问题描述
今天,我有一个按地点分组的个人列表.我想产生一个新变量,根据每个人的位置给每个人一个数字.我希望数据看起来像这样:
G'day, I have a list of individuals that are grouped by place. I want to produce a new variable that gives a number to each individual dependant on their place. What I would like my data to look like is:
place individual
here 1
here 2
here 3
there 1
there 2
somewhere 1
somewhere 2
我写了这个:
nest="ddd", "ddd", "fff", "fff", "fff", "fff", "qqq", "qqq"
def individual(x):
i = 0
j = 1
while i < len(x):
if x[i] == x[i-1]:
print(j+1)
i = i + 1
j = j + 1
else:
print(1)
i = i + 1
j = 1
individual(nest)
这会打印出我想要的值,但是,当我在其中放回车符时,它会跳出循环并仅返回第一个值.我想知道如何返回这些值,以便将它们作为新列添加到我的数据中?
This prints out the values I want, however, when I put return in there it breaks out of the loop and only returns the first value. I was wondering how I could return these values, so that I can add them to my data as a new column?
我了解到产量吗?但不确定是否合适.谢谢您的帮助!
I read about yield? but was unsure if it is appropriate. Thank you for your help!
干杯,亚当
推荐答案
将print(...)
替换为yield ...
.那么您将拥有一个生成器,它将为您提供迭代.然后,可以通过遍历结果将其转换为其他适当的数据结构.例如,要从生成器构建列表,您可以执行以下操作:
replace print(...)
with yield ...
. then you'll have a generator which will give you an iterable. You can then turn that into some other appropriate data-structure by iterating over the result. For example, to construct a list from the generator, you could do:
list(individual(nest)) #this is prefered
在这种情况下,迭代是隐式的...
Where the iteration is implicit in this case ...
或(在这种情况下更全面,但可能提供更多信息):
or (the more round-about but possibly more informative in this context):
[ x for x in individual(nest) ] #This is just for demonstration.
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