在元组列表中查找元素

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

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

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