问题：在元组列表中查找元素
我有一个清单“ a”
a= [(1,2),(1,4),(3,5),(5,7)]
我需要找到一个特定数字的所有元组。说1
result = [(1,2),(1,4)]
我怎么做？
I have a list ‘a’
a= [(1,2),(1,4),(3,5),(5,7)]
I need to find all the tuples for a particular number. say for 1 it will be
result = [(1,2),(1,4)]
How do I do that?
回答 0
如果只希望第一个数字匹配，则可以这样操作：
[item for item in a if item[0] == 1]
如果您仅搜索其中包含1的元组：
[item for item in a if 1 in item]
If you just want the first number to match you can do it like this:
[item for item in a if item[0] == 1]
If you are just searching for tuples with 1 in them:
[item for item in a if 1 in item]
回答 1
实际上，有一种聪明的方法可以用于任何元组列表，其中每个元组的大小为2：您可以将列表转换成一个字典。
例如，
test = [("hi", 1), ("there", 2)]
test = dict(test)
print test["hi"] # prints 1
There is actually a clever way to do this that is useful for any list of tuples where the size of each tuple is 2: you can convert your list into a single dictionary.
For example,
test = [("hi", 1), ("there", 2)]
test = dict(test)
print test["hi"] # prints 1
回答 2
阅读列表理解
[ (x,y) for x, y in a if x == 1 ]
还要阅读生成器函数和yield语句。
def filter_value( someList, value ):
for x, y in someList:
if x == value :
yield x,y
result= list( filter_value( a, 1 ) )
Read up on List Comprehensions
[ (x,y) for x, y in a if x == 1 ]
Also read up up generator functions and the yield statement.
def filter_value( someList, value ):
for x, y in someList:
if x == value :
yield x,y
result= list( filter_value( a, 1 ) )
回答 3
[tup for tup in a if tup[0] == 1]
[tup for tup in a if tup[0] == 1]
回答 4
for item in a:
if 1 in item:
print item
for item in a:
if 1 in item:
print item
回答 5
>>> [i for i in a if 1 in i]
[（1，2），（1，4）]
>>> [i for i in a if 1 in i]
[(1, 2), (1, 4)]
回答 6
该filter函数还可以提供一个有趣的解决方案：
result = list(filter(lambda x: x.count(1) > 0, a))
它会在列表中的元组中搜索是否出现1。如果搜索仅限于第一个元素，则可以将解决方案修改为：
result = list(filter(lambda x: x[0] == 1, a))
The filter function can also provide an interesting solution:
result = list(filter(lambda x: x.count(1) > 0, a))
which searches the tuples in the list a for any occurrences of 1. If the search is limited to the first element, the solution can be modified into:
result = list(filter(lambda x: x[0] == 1, a))
回答 7
使用过滤功能：
>>> def get_values（iterables，key_to_find）：
返回列表（过滤器（lambda x：x中的key_to_find，可迭代））
>>> a = [（1,2 ,,（1,4），（3,5），（5,7）]
>>> get_values（a，1）
>>> [（1，2），（1，4）]
Using filter function:
>>> def get_values(iterables, key_to_find):
return list(filter(lambda x:key_to_find in x, iterables))
>>> a = [(1,2),(1,4),(3,5),(5,7)]
>>> get_values(a, 1)
>>> [(1, 2), (1, 4)]
回答 8
或takewhile，（此外，还会显示更多值的示例）：
>>> a= [(1,2),(1,4),(3,5),(5,7),(0,2)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,a))
[(1, 2), (1, 4)]
>>>
如果未排序，例如：
>>> a= [(1,2),(3,5),(1,4),(5,7)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,sorted(a,key=lambda x: x[0]==1)))
[(1, 2), (1, 4)]
>>>
Or takewhile, ( addition to this, example of more values is shown ):
>>> a= [(1,2),(1,4),(3,5),(5,7),(0,2)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,a))
[(1, 2), (1, 4)]
>>>
if unsorted, like:
>>> a= [(1,2),(3,5),(1,4),(5,7)]
>>> import itertools
>>> list(itertools.takewhile(lambda x: x[0]==1,sorted(a,key=lambda x: x[0]==1)))
[(1, 2), (1, 4)]
>>>
回答 9
如果要在元组中搜索元组中存在的任何数字，则可以使用
a= [(1,2),(1,4),(3,5),(5,7)]
i=1
result=[]
for j in a:
if i in j:
result.append(j)
print(result)
if i==j[0] or i==j[index]如果要搜索特定索引中的数字，也可以使用
if you want to search tuple for any number which is present in tuple then you can use
a= [(1,2),(1,4),(3,5),(5,7)]
i=1
result=[]
for j in a:
if i in j:
result.append(j)
print(result)
You can also use if i==j[0] or i==j[index] if you want to search a number in particular index
