如何在Python中搜索元组列表

Question

So I have a list of tuples such as this:

[(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]

I want this list for a tuple whose number value is equal to something.

So that if I do search(53) it will return the index value of 2

Is there an easy way to do this?

Answer 1

[i for i, v in enumerate(L) if v[0] == 53]

Answer 2

You can use a list comprehension :

>>> a = [(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]
>>> [x[0] for x in a]
[1, 22, 53, 44]
>>> [x[0] for x in a].index(53)
2

Answer 3

tl;dr

A generator expression is probably the most performant and simple solution to your problem:

l = [(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]

result = next((i for i, v in enumerate(l) if v[0] == 53), None)
# 2

Explanation

There are several answers that provide a simple solution to this question with list comprehensions. While these answers are perfectly correct, they are not optimal. Depending on your use case, there may be significant benefits to making a few simple modifications.

The main problem I see with using a list comprehension for this use case is that the entire list will be processed, although you only want to find 1 element .

Python provides a simple construct which is ideal here. It is called the generator expression . Here is an example:

# Our input list, same as before
l = [(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]

# Call next on our generator expression.
next((i for i, v in enumerate(l) if v[0] == 53), None)

We can expect this method to perform basically the same as list comprehensions in our trivial example, but what if we're working with a larger data set? That's where the advantage of using the generator method comes into play. Rather than constructing a new list, we'll use your existing list as our iterable, and use next() to get the first item from our generator.

Lets look at how these methods perform differently on some larger data sets. These are large lists, made of 10000000 + 1 elements, with our target at the beginning (best) or end (worst). We can verify that both of these lists will perform equally using the following list comprehension:

List comprehensions

"Worst case"

worst_case = ([(False, 'F')] * 10000000) + [(True, 'T')]
print [i for i, v in enumerate(worst_case) if v[0] is True]

# [10000000]
#          2 function calls in 3.885 seconds
#
#    Ordered by: standard name
#
#    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
#         1    3.885    3.885    3.885    3.885 so_lc.py:1(<module>)
#         1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

"Best case"

best_case = [(True, 'T')] + ([(False, 'F')] * 10000000)
print [i for i, v in enumerate(best_case) if v[0] is True]

# [0]
#          2 function calls in 3.864 seconds
#
#    Ordered by: standard name
#
#    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
#         1    3.864    3.864    3.864    3.864 so_lc.py:1(<module>)
#         1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}

Generator expressions

Here's my hypothesis for generators: we'll see that generators will significantly perform better in the best case, but similarly in the worst case. This performance gain is mostly due to the fact that the generator is evaluated lazily, meaning it will only compute what is required to yield a value.

Worst case

# 10000000
#          5 function calls in 1.733 seconds
#
#    Ordered by: standard name
#
#    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
#         2    1.455    0.727    1.455    0.727 so_lc.py:10(<genexpr>)
#         1    0.278    0.278    1.733    1.733 so_lc.py:9(<module>)
#         1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
#         1    0.000    0.000    1.455    1.455 {next}

Best case

best_case  = [(True, 'T')] + ([(False, 'F')] * 10000000)
print next((i for i, v in enumerate(best_case) if v[0] == True), None)

# 0
#          5 function calls in 0.316 seconds
#
#    Ordered by: standard name
#
#    ncalls  tottime  percall  cumtime  percall filename:lineno(function)
#         1    0.316    0.316    0.316    0.316 so_lc.py:6(<module>)
#         2    0.000    0.000    0.000    0.000 so_lc.py:7(<genexpr>)
#         1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}
#         1    0.000    0.000    0.000    0.000 {next}

WHAT?! The best case blows away the list comprehensions, but I wasn't expecting the our worst case to outperform the list comprehensions to such an extent. How is that? Frankly, I could only speculate without further research.

Take all of this with a grain of salt, I have not run any robust profiling here, just some very basic testing. This should be sufficient to appreciate that a generator expression is more performant for this type of list searching.

Note that this is all basic, built-in python. We don't need to import anything or use any libraries.

I first saw this technique for searching in the Udacity cs212 course with Peter Norvig.

Answer 4

Your tuples are basically key-value pairs--a python dict --so:

l = [(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]
val = dict(l)[53]

Edit -- aha, you say you want the index value of (53, "xuxa"). If this is really what you want, you'll have to iterate through the original list, or perhaps make a more complicated dictionary:

d = dict((n,i) for (i,n) in enumerate(e[0] for e in l))
idx = d[53]

Answer 5

Hmm... well, the simple way that comes to mind is to convert it to a dict

d = dict(thelist)

and access d[53] .

EDIT : Oops, misread your question the first time. It sounds like you actually want to get the index where a given number is stored. In that case, try

dict((t[0], i) for i, t in enumerate(thelist))

instead of a plain old dict conversion. Then d[53] would be 2.

Answer 6

Supposing the list may be long and the numbers may repeat, consider using the SortedList type from the Python sortedcontainers module . The SortedList type will automatically maintain the tuples in order by number and allow for fast searching.

For example:

from sortedcontainers import SortedList
sl = SortedList([(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")])

# Get the index of 53:

index = sl.bisect((53,))

# With the index, get the tuple:

tup = sl[index]

This will work a lot faster than the list comprehension suggestion by doing a binary search. The dictionary suggestion will be faster still but won't work if there could be duplicate numbers with different strings.

If there are duplicate numbers with different strings then you need to take one more step:

end = sl.bisect((53 + 1,))

results = sl[index:end]

By bisecting for 54, we will find the end index for our slice. This will be significantly faster on long lists as compared with the accepted answer.

Answer 7

只是另一种方式。

zip(*a)[0].index(53)

Answer 8

[k for k,v in l if v ==' delicia ']

here l is the list of tuples-[(1,"juca"),(22,"james"),(53,"xuxa"),(44,"delicia")]

And instead of converting it to a dict, we are using llist comprehension.

*Key* in Key,Value in list, where value = **delicia**