在元组列表中查找元素

I have a list 'a'

我有一个列表'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

我需要找到特定数字的所有元组。说它会是1

result = [(1,2),(1,4)]

How do I do that?

我怎么做？

8 个解决方案

#1

162

If you just want the first number to match you can do it like this:

如果您只想匹配第一个数字，可以这样做：

[item for item in a if item[0] == 1]

If you are just searching for tuples with 1 in them:

如果您只是搜索其中包含1的元组：

[item for item in a if 1 in item]

#2

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.

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：您可以将列表转换为单个字典。

For example,

例如，

test = [("hi", 1), ("there", 2)]
test = dict(test)
print test["hi"] # prints 1

#3

Read up on List Comprehensions

阅读列表理解

[ (x,y) for x, y in a if x  == 1 ]

Also read up up generator functions and the yield statement.

还要读取生成器函数和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 ) )

#4

[tup for tup in a if tup[0] == 1]

#5

for item in a:
   if 1 in item:
       print item

#6

>>> [i for i in a if 1 in i]

[(1, 2), (1, 4)]

[（1,2），（1,4）]

#7

The filter function can also provide an interesting solution:

过滤功能还可以提供一个有趣的解决方案：

result = list(filter(lambda x: x.count(1) > 0, a))

which searches the tuples in list for any occurrence of 1. If the search is limited to the first element, the solution can be modified into:

在列表中搜索任何出现的1的元组。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))

#8

Using filter function:

使用过滤功能：

>>> 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)]

#1

162