[英]Trying to return relevant information from a list of lists based on user input
[英]Return lists from a list of lists when given an input value
我已经在这个问题上坚持了一段时间,但似乎无法找到解决方案,因此我请求帮助。 PS:编程还是有点新鲜
我在列表中有列表:
[(2012, 'january', 'monday'), (2012, 'february', 'monday'), (2012, 'january', 'tuesday')]
我想要的是一个带有列表的新列表,当输入“monday”时:
[(2012, 'january', 'monday'), (2012, 'february', 'monday')]
到目前为止我的代码:
lists = [(2012, 'january', 'monday'), (2012, 'february', 'monday'), (2012, 'january', 'tuesday')]
day = input("Give day: ") #monday
def select_monday(lists, day):
list2 = []
for list in lists:
if list[2] == day: #from here I'm stuck and do not know how to continue
list2.append(list[2])
else:
return None
return list2
结果: None
我不知道如何获得具有特定值的所有列表
你的代码很好,除了你不需要else
语句,因为否则在下一次迭代中你会丢失前面步骤的结果; 你也应该实际调用你的函数:
lists = [(2012, 'january', 'monday'), (2012, 'february', 'monday'), (2012, 'january', 'tuesday')]
day = input("Give day: ") #monday
def select_monday(lists, day):
list2 = []
for list in lists:
if list[2] == day: #from here I'm stuck and do not know how to continue
list2.append(list)
return list2
print(select_monday(lists, day))
这是一个更紧凑的功能:
def select_monday_2(lists, day):
return list(filter(lambda x: x[2] == day, lists))
print(select_monday_2(lists, day))
如果当天不是星期一,则返回“ None
”。 这效果更好:
def select_monday(lists, day):
list2 = []
for lst in lists:
if lst[2] == day:
list2.append(lst)
return list2
此外,附上整个清单,而不仅仅是工作日。 最后,最好不要使用list
作为变量名,因为它是内置的。
现在:
>>> select_monday(lists, day)
[(2012, 'january', 'monday'), (2012, 'february', 'monday')]
更短的替代方案,使用列表理解:
>>> [x for x in lists if x[2] == day]
[(2012, 'january', 'monday'), (2012, 'february', 'monday')]
这个问题是else语句。 如果其中一个检查是假的,那么它将return None
。 删除else语句并删除append
语句中列表中的[2]
,除非您只想追加当天。
lists = [(2012, 'january', 'monday'), (2012, 'february', 'monday'), (2012, 'january', 'tuesday')]
day = input("Give day: ") #monday
def select_monday(lists, day):
list2 = []
for list in lists:
if list[2] == day: #from here I'm stuck and do not know how to continue
list2.append(list)
return list2
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