Asked  6 Months ago    Answers:  5   Viewed   45 times

I have this nested list:

l = [['40', '20', '10', '30'], ['20', '20', '20', '20', '20', '30', '20'], ['30', '20', '30', '50', '10', '30', '20', '20', '20'], ['100', '100'], ['100', '100', '100', '100', '100'], ['100', '100', '100', '100']]

Now, what I want to do is convert each element in a list to float. My solution is this:

newList = []
for x in l:
  for y in x:
    newList.append(float(y))

But can this be done using nested list comprehension, right?

what I've done is:

[float(y) for y in x for x in l]

But then the result is bunch of 100's with the sum of 2400.

any solution, an explanation would be much appreciated. Thanks!

 Answers

19

Here is how you would do this with a nested list comprehension:

[[float(y) for y in x] for x in l]

This would give you a list of lists, similar to what you started with except with floats instead of strings. If you want one flat list then you would use [float(y) for x in l for y in x].

Tuesday, June 1, 2021
 
TuomasR
answered 6 Months ago
54

Assuming you are using Python 3.x:

print(*myList, sep='n')

You can get the same behavior on Python 2.x using from __future__ import print_function, as noted by mgilson in comments.

With the print statement on Python 2.x you will need iteration of some kind, regarding your question about print(p) for p in myList not working, you can just use the following which does the same thing and is still one line:

for p in myList: print p

For a solution that uses 'n'.join(), I prefer list comprehensions and generators over map() so I would probably use the following:

print 'n'.join(str(p) for p in myList) 
Tuesday, June 1, 2021
 
simPod
answered 6 Months ago
46

How are lists and other sequences compared in Python?

Lists (and other sequences) in Python are compared lexicographically and not based on any other parameter.

Sequence objects may be compared to other objects with the same sequence type. The comparison uses lexicographical ordering: first the first two items are compared, and if they differ this determines the outcome of the comparison; if they are equal, the next two items are compared, and so on, until either sequence is exhausted.


What is lexicographic sorting?

From the Wikipedia page on lexicographic sorting

lexicographic or lexicographical order (also known as lexical order, dictionary order, alphabetical order or lexicographic(al) product) is a generalization of the way the alphabetical order of words is based on the alphabetical order of their component letters.

The min function returns the smallest value in the iterable. So the lexicographic value of [1,2] is the least in that list. You can check by using [1,2,21]

>>> my_list=[[1,2,21],[1,3],[1,2]]
>>> min(my_list)
[1, 2]

What is happening in this case of min?

Going element wise on my_list, firstly [1,2,21] and [1,3]. Now from the docs

If two items to be compared are themselves sequences of the same type, the lexicographical comparison is carried out recursively.

Thus the value of [1,1,21] is less than [1,3], because the second element of [1,3], which is, 3 is lexicographically higher than the value of the second element of [1,1,21], which is, 1.

Now comparing [1,2] and [1,2,21], and adding another reference from the docs

If one sequence is an initial sub-sequence of the other, the shorter sequence is the smaller (lesser) one.

[1,2] is an initial sub-sequence of [1,2,21]. Therefore the value of [1,2] on the whole is smaller than that of [1,2,21]. Hence [1,2] is returned as the output.

This can be validated by using the sorted function

>>> sorted(my_list)
[[1, 2], [1, 2, 21], [1, 3]]

What if the list has multiple minimum elements?

If the list contains duplicate min elements the first is returned

>>> my_list=[[1,2],[1,2]]
>>> min(my_list)
[1, 2]

This can be confirmed using the id function call

>>> my_list=[[1,2],[1,2]]
>>> [id(i) for i in my_list]
[140297364849368, 140297364850160]
>>> id(min(my_list))
140297364849368

What do I need to do to prevent lexicographic comparison in min?

If the required comparison is not lexicographic then the key argument can be used (as mentioned by Padraic)

The min function has an additional optional argument called key. The key argument takes a function.

The optional key argument specifies a one-argument ordering function like that used for list.sort(). The key argument, if supplied, must be in keyword form (for example, min(a,b,c,key=func)).

For example, if we need the smallest element by length, we need to use the len function.

>>> my_list=[[1,2,21],[1,3],[1,2]]
>>> min(my_list,key=len)            # Notice the key argument
[1, 3]

As we can see the first shortest element is returned here.


What if the list is heterogeneous?

Until Python2

If the list is heterogeneous type names are considered for ordering, check Comparisions,

Objects of different types except numbers are ordered by their type names

Hence if you put an int and a list there you will get the integer value as the smallest as i is of lower value than l. Similarly '1' would be of higher value than both of this.

>>> my_list=[[1,1,21],1,'1']
>>> min(my_list)
1

Python3 and onwards

However this confusing technique was removed in Python3. It now raises a TypeError. Read What's new in Python 3.0

The ordering comparison operators (<, <=, >=, >) raise a TypeError exception when the operands don’t have a meaningful natural ordering. Thus, expressions like 1 < '', 0 > None or len <= len are no longer valid, and e.g. None < None raises TypeError instead of returning False. A corollary is that sorting a heterogeneous list no longer makes sense – all the elements must be comparable to each other.

>>> my_list=[[1,1,21],1,'1']
>>> min(my_list)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: int() < list()

But it works for Comparable types, For example

>>> my_list=[1,2.0]
>>> min(my_list)
1

Here we can see that the list contains float values and int values. But as float and int are comparable types, min function works in this case.

Wednesday, June 23, 2021
 
ramdemon
answered 6 Months ago
66

The list comprehension executes the loop in Python bytecode, just like a regular for loop.

The list() call iterates entirely in C code, which is far faster.

The bytecode for the list comprehension looks like this:

>>> import dis
>>> dis.dis(compile("[x for x in xrange(1000000)]", '<stdin>', 'exec'))
  1           0 BUILD_LIST               0
              3 LOAD_NAME                0 (xrange)
              6 LOAD_CONST               0 (1000000)
              9 CALL_FUNCTION            1
             12 GET_ITER            
        >>   13 FOR_ITER                12 (to 28)
             16 STORE_NAME               1 (x)
             19 LOAD_NAME                1 (x)
             22 LIST_APPEND              2
             25 JUMP_ABSOLUTE           13
        >>   28 POP_TOP             
             29 LOAD_CONST               1 (None)
             32 RETURN_VALUE        

The >> pointers roughly give you the boundaries of the loop being executed, so you have 1 million STORE_NAME, LOAD_NAME and LIST_APPEND steps to execute in the Python bytecode evaluation loop.

list() on the other hand just grabs the values from the xrange() iterable directly by using the C API for object iteration, and it can use the length of the xrange() object to pre-allocate the list object rather than grow it dynamically.

Friday, July 23, 2021
 
Ticksy
answered 5 Months ago
52

While the hierarchy you posted above really doesn't make much sense to me (seems RoomType and Room are backwards), I'll post an example to go with it:

Hotels.SelectMany(h => h.RoomType)
      .Select(rt => rt.Room.Id)
      .Distinct()
      .ToArray();
Wednesday, September 8, 2021
 
pirtle
answered 3 Months ago
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