Related
I would like to know if there is a better way to print all objects in a Python list than this :
myList = [Person("Foo"), Person("Bar")]
print("\n".join(map(str, myList)))
Foo
Bar
I read this way is not really good :
myList = [Person("Foo"), Person("Bar")]
for p in myList:
print(p)
Isn't there something like :
print(p) for p in myList
If not, my question is... why ? If we can do this kind of stuff with comprehensive lists, why not as a simple statement outside a list ?
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)
I use this all the time :
#!/usr/bin/python
l = [1,2,3,7]
print "".join([str(x) for x in l])
[print(a) for a in list] will give a bunch of None types at the end though it prints out all the items
For Python 2.*:
If you overload the function __str__() for your Person class, you can omit the part with map(str, ...). Another way for this is creating a function, just like you wrote:
def write_list(lst):
for item in lst:
print str(item)
...
write_list(MyList)
There is in Python 3.* the argument sep for the print() function. Take a look at documentation.
Expanding #lucasg's answer (inspired by the comment it received):
To get a formatted list output, you can do something along these lines:
l = [1,2,5]
print ", ".join('%02d'%x for x in l)
01, 02, 05
Now the ", " provides the separator (only between items, not at the end) and the formatting string '02d'combined with %x gives a formatted string for each item x - in this case, formatted as an integer with two digits, left-filled with zeros.
To display each content, I use:
mylist = ['foo', 'bar']
indexval = 0
for i in range(len(mylist)):
print(mylist[indexval])
indexval += 1
Example of using in a function:
def showAll(listname, startat):
indexval = startat
try:
for i in range(len(mylist)):
print(mylist[indexval])
indexval = indexval + 1
except IndexError:
print('That index value you gave is out of range.')
Hope I helped.
I think this is the most convenient if you just want to see the content in the list:
myList = ['foo', 'bar']
print('myList is %s' % str(myList))
Simple, easy to read and can be used together with format string.
I recently made a password generator and although I'm VERY NEW to python, I whipped this up as a way to display all items in a list (with small edits to fit your needs...
x = 0
up = 0
passwordText = ""
password = []
userInput = int(input("Enter how many characters you want your password to be: "))
print("\n\n\n") # spacing
while x <= (userInput - 1): #loops as many times as the user inputs above
password.extend([choice(groups.characters)]) #adds random character from groups file that has all lower/uppercase letters and all numbers
x = x+1 #adds 1 to x w/o using x ++1 as I get many errors w/ that
passwordText = passwordText + password[up]
up = up+1 # same as x increase
print(passwordText)
Like I said, IM VERY NEW to Python and I'm sure this is way to clunky for a expert, but I'm just here for another example
Assuming you are fine with your list being printed [1,2,3], then an easy way in Python3 is:
mylist=[1,2,3,'lorem','ipsum','dolor','sit','amet']
print(f"There are {len(mylist):d} items in this lorem list: {str(mylist):s}")
Running this produces the following output:
There are 8 items in this lorem list: [1, 2, 3, 'lorem', 'ipsum',
'dolor', 'sit', 'amet']
OP's question is: does something like following exists, if not then why
print(p) for p in myList # doesn't work, OP's intuition
answer is, it does exist which is:
[p for p in myList] #works perfectly
Basically, use [] for list comprehension and get rid of print to avoiding printing None. To see why print prints None see this
To print each element of a given list using a single line code
for i in result: print(i)
You can also make use of the len() function and identify the length of the list to print the elements as shown in the below example:
sample_list = ['Python', 'is', 'Easy']
for i in range(0, len(sample_list)):
print(sample_list[i])
Reference : https://favtutor.com/blogs/print-list-python
you can try doing this: this will also print it as a string
print(''.join([p for p in myList]))
or if you want to a make it print a newline every time it prints something
print(''.join([p+'\n' for p in myList]))
So my Python code to keep on finding the sum of a number's digits (9983 = 9+9+8+3 = 29 = 2+9 = 11 = 1+1 = 2) below isn't working, and it gives me this exception
ValueError: invalid literal for int() with base 10: '['".
It was working before when instead of using list comprehension, I did it by iterating through the list normally, so I suspect it's a problem with the list comprehension? Help?
x = int(input())
def rec(x):
if (x<10):
return x
return rec(sum(x))
def sum(x):
lst = [int(y) for y in list(str(x))]
return sum(lst)
print(rec(x))
There's a built-in function called sum, and you defined a function with the same name - but you need the original function as part of the solution.
Let's start by renaming your version of sum to summation, and let's take the opportunity to simplify the code a bit more:
def rec(x):
if x < 10:
return x
return rec(summation(x))
def summation(x):
return sum(int(y) for y in str(x))
See how the built-in sum was needed inside summation? that's why your function must have a different name! And although we could have used a list comprehension, it's easier and more efficient to pass a generator expression to sum. Alternatively (but less efficiently) we could have used a list comprehension:
def summation(x):
lst = [int(y) for y in str(x)]
return sum(lst)
Either way, it works as expected:
rec(9983)
=>2
You are creating confusion for Python ;-)
The problem is in def sum(x):, not in the list comprehension. sum is a reserved keyword. Just rename the function (and the call to the function) and it works.
I would like to know if there is a better way to print all objects in a Python list than this :
myList = [Person("Foo"), Person("Bar")]
print("\n".join(map(str, myList)))
Foo
Bar
I read this way is not really good :
myList = [Person("Foo"), Person("Bar")]
for p in myList:
print(p)
Isn't there something like :
print(p) for p in myList
If not, my question is... why ? If we can do this kind of stuff with comprehensive lists, why not as a simple statement outside a list ?
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)
I use this all the time :
#!/usr/bin/python
l = [1,2,3,7]
print "".join([str(x) for x in l])
[print(a) for a in list] will give a bunch of None types at the end though it prints out all the items
For Python 2.*:
If you overload the function __str__() for your Person class, you can omit the part with map(str, ...). Another way for this is creating a function, just like you wrote:
def write_list(lst):
for item in lst:
print str(item)
...
write_list(MyList)
There is in Python 3.* the argument sep for the print() function. Take a look at documentation.
Expanding #lucasg's answer (inspired by the comment it received):
To get a formatted list output, you can do something along these lines:
l = [1,2,5]
print ", ".join('%02d'%x for x in l)
01, 02, 05
Now the ", " provides the separator (only between items, not at the end) and the formatting string '02d'combined with %x gives a formatted string for each item x - in this case, formatted as an integer with two digits, left-filled with zeros.
To display each content, I use:
mylist = ['foo', 'bar']
indexval = 0
for i in range(len(mylist)):
print(mylist[indexval])
indexval += 1
Example of using in a function:
def showAll(listname, startat):
indexval = startat
try:
for i in range(len(mylist)):
print(mylist[indexval])
indexval = indexval + 1
except IndexError:
print('That index value you gave is out of range.')
Hope I helped.
I think this is the most convenient if you just want to see the content in the list:
myList = ['foo', 'bar']
print('myList is %s' % str(myList))
Simple, easy to read and can be used together with format string.
I recently made a password generator and although I'm VERY NEW to python, I whipped this up as a way to display all items in a list (with small edits to fit your needs...
x = 0
up = 0
passwordText = ""
password = []
userInput = int(input("Enter how many characters you want your password to be: "))
print("\n\n\n") # spacing
while x <= (userInput - 1): #loops as many times as the user inputs above
password.extend([choice(groups.characters)]) #adds random character from groups file that has all lower/uppercase letters and all numbers
x = x+1 #adds 1 to x w/o using x ++1 as I get many errors w/ that
passwordText = passwordText + password[up]
up = up+1 # same as x increase
print(passwordText)
Like I said, IM VERY NEW to Python and I'm sure this is way to clunky for a expert, but I'm just here for another example
Assuming you are fine with your list being printed [1,2,3], then an easy way in Python3 is:
mylist=[1,2,3,'lorem','ipsum','dolor','sit','amet']
print(f"There are {len(mylist):d} items in this lorem list: {str(mylist):s}")
Running this produces the following output:
There are 8 items in this lorem list: [1, 2, 3, 'lorem', 'ipsum',
'dolor', 'sit', 'amet']
OP's question is: does something like following exists, if not then why
print(p) for p in myList # doesn't work, OP's intuition
answer is, it does exist which is:
[p for p in myList] #works perfectly
Basically, use [] for list comprehension and get rid of print to avoiding printing None. To see why print prints None see this
To print each element of a given list using a single line code
for i in result: print(i)
You can also make use of the len() function and identify the length of the list to print the elements as shown in the below example:
sample_list = ['Python', 'is', 'Easy']
for i in range(0, len(sample_list)):
print(sample_list[i])
Reference : https://favtutor.com/blogs/print-list-python
you can try doing this: this will also print it as a string
print(''.join([p for p in myList]))
or if you want to a make it print a newline every time it prints something
print(''.join([p+'\n' for p in myList]))
I have done some writing like below
new_list = ["000","122","121","230"]
for item in new_list:
r = [np.matrix([[0.5,0.5],[0,1]]),
np.matrix([[0.5,0.25],[0,1]]),
np.matrix([[0.5,0.25],[0,1]]),
np.matrix([[0.5,0],[0,1]])]
(W0,W1,W2,W3) = r
for i in item:
i == 0
if item[i] == 0:
return W0
elif item[i] == 1:
return W1
elif item[i] == 2:
return W2
else:
return W3
i = i + 1
list_new = [np.matrix([1,0]) * new_list * np.matrix([0.5],[1])]
print(list_new)
I want to create a program for example when the input from the list is "122", it will multiply each element inside "122" as matrix W1*W2*W2.
The final result should be like
np.matrix([1,0]) * new_list2 * np.matrix([0.5],[1])
The
new_list2
here should be all multiplication result from the new_list. Meaning
new_list2 = [W0*W0*W0,W1*W2*W2,...]
I run this code and keep getting a syntax error saying
"return" outside function
I want to ask did I missed something here or there is other function/way to make it done using python. I am using Python 3.5
Thank you for the answer.
As the error message says, you are using "return" outside a function, which just doesn't make sense since "return" means "exit the function now and pass this value (or None if you don't specify a value) to the caller". If you're not inside function there is no caller to returns to...
Something like this?
import numpy as np
from functools import reduce
new_list = ["000", "122", "121", "230"]
matrices = [
np.matrix([[0.5,0.50],[0,1]]),
np.matrix([[0.5,0.25],[0,1]]),
np.matrix([[0.5,0.25],[0,1]]),
np.matrix([[0.5,0.00],[0,1]])
]
def multiply(s):
mt_seq = (matrices[i] for i in map(int, item))
return reduce(np.multiply, mt_seq) # or reversed(mt_seq)
for item in new_list:
print(multiply(item))
You can just convert each digit in the string with int(digit), e.g. int('2'), and do it for every element of the string with map(int, '01234').
Then, it's just a matter of taking those matrices and multiplying them. (Sorry, I am not familiar with np.multiply and np.matrix and maybe you need to reverse the order of multplication)
reduce(fun, seq) applies fun to seq, like this: fun(fun(fun(seq[0], seq[1]), seq[2]), ...)
I'm calling a bunch of methods that return a list. The list may be empty. If the list is non-empty, I want to return the first item; otherwise, I want to return None. This code works:
def main():
my_list = get_list()
if len(my_list) > 0:
return my_list[0]
return None
but it seems to me that there should be a simple one-line idiom for doing this. Is there?
Python 2.6+
next(iter(your_list), None)
If your_list can be None:
next(iter(your_list or []), None)
Python 2.4
def get_first(iterable, default=None):
if iterable:
for item in iterable:
return item
return default
Example:
x = get_first(get_first_list())
if x:
...
y = get_first(get_second_list())
if y:
...
Another option is to inline the above function:
for x in get_first_list() or []:
# process x
break # process at most one item
for y in get_second_list() or []:
# process y
break
To avoid break you could write:
for x in yield_first(get_first_list()):
x # process x
for y in yield_first(get_second_list()):
y # process y
Where:
def yield_first(iterable):
for item in iterable or []:
yield item
return
The best way is this:
a = get_list()
return a[0] if a else None
You could also do it in one line, but it's much harder for the programmer to read:
return (get_list()[:1] or [None])[0]
(get_list() or [None])[0]
That should work.
BTW I didn't use the variable list, because that overwrites the builtin list() function.
The most python idiomatic way is to use the next() on a iterator since list is iterable. just like what #J.F.Sebastian put in the comment on Dec 13, 2011.
next(iter(the_list), None) This returns None if the_list is empty. see next() Python 2.6+
or if you know for sure the_list is not empty:
iter(the_list).next() see iterator.next() Python 2.2+
If you find yourself trying to pluck the first thing (or None) from a list comprehension you can switch to a generator to do it like:
next((x for x in blah if cond), None)
Pro: works if blah isn't indexable Con: it's unfamiliar syntax. It's useful while hacking around and filtering stuff in ipython though.
The OP's solution is nearly there, there are just a few things to make it more Pythonic.
For one, there's no need to get the length of the list. Empty lists in Python evaluate to False in an if check. Just simply say
if list:
Additionally, it's a very Bad Idea to assign to variables that overlap with reserved words. "list" is a reserved word in Python.
So let's change that to
some_list = get_list()
if some_list:
A really important point that a lot of solutions here miss is that all Python functions/methods return None by default. Try the following below.
def does_nothing():
pass
foo = does_nothing()
print foo
Unless you need to return None to terminate a function early, it's unnecessary to explicitly return None. Quite succinctly, just return the first entry, should it exist.
some_list = get_list()
if some_list:
return list[0]
And finally, perhaps this was implied, but just to be explicit (because explicit is better than implicit), you should not have your function get the list from another function; just pass it in as a parameter. So, the final result would be
def get_first_item(some_list):
if some_list:
return list[0]
my_list = get_list()
first_item = get_first_item(my_list)
As I said, the OP was nearly there, and just a few touches give it the Python flavor you're looking for.
Python idiom to return first item or None?
The most Pythonic approach is what the most upvoted answer demonstrated, and it was the first thing to come to my mind when I read the question. Here's how to use it, first if the possibly empty list is passed into a function:
def get_first(l):
return l[0] if l else None
And if the list is returned from a get_list function:
l = get_list()
return l[0] if l else None
New in Python 3.8, Assignment Expressions
Assignment expressions use the in-place assignment operator (informally called the walrus operator), :=, new in Python 3.8, allows us to do the check and assignment in-place, allowing the one-liner:
return l[0] if (l := get_list()) else None
As a long-time Python user, this feels like we're trying to do too much on one line - I feel it would be better style to do the presumptively equally performant:
if l := get_list():
return l[0]
return None
In support of this formulation is Tim Peter's essay in the PEP proposing this change to the language. He didn't address the first formulation, but based on the other formulations he did like, I don't think he would mind.
Other ways demonstrated to do this here, with explanations
for
When I began trying to think of clever ways to do this, this is the second thing I thought of:
for item in get_list():
return item
This presumes the function ends here, implicitly returning None if get_list returns an empty list. The below explicit code is exactly equivalent:
for item in get_list():
return item
return None
if some_list
The following was also proposed (I corrected the incorrect variable name) which also uses the implicit None. This would be preferable to the above, as it uses the logical check instead of an iteration that may not happen. This should be easier to understand immediately what is happening. But if we're writing for readability and maintainability, we should also add the explicit return None at the end:
some_list = get_list()
if some_list:
return some_list[0]
slice or [None] and select zeroth index
This one is also in the most up-voted answer:
return (get_list()[:1] or [None])[0]
The slice is unnecessary, and creates an extra one-item list in memory. The following should be more performant. To explain, or returns the second element if the first is False in a boolean context, so if get_list returns an empty list, the expression contained in the parentheses will return a list with 'None', which will then be accessed by the 0 index:
return (get_list() or [None])[0]
The next one uses the fact that and returns the second item if the first is True in a boolean context, and since it references my_list twice, it is no better than the ternary expression (and technically not a one-liner):
my_list = get_list()
return (my_list and my_list[0]) or None
next
Then we have the following clever use of the builtin next and iter
return next(iter(get_list()), None)
To explain, iter returns an iterator with a .next method. (.__next__ in Python 3.) Then the builtin next calls that .next method, and if the iterator is exhausted, returns the default we give, None.
redundant ternary expression (a if b else c) and circling back
The below was proposed, but the inverse would be preferable, as logic is usually better understood in the positive instead of the negative. Since get_list is called twice, unless the result is memoized in some way, this would perform poorly:
return None if not get_list() else get_list()[0]
The better inverse:
return get_list()[0] if get_list() else None
Even better, use a local variable so that get_list is only called one time, and you have the recommended Pythonic solution first discussed:
l = get_list()
return l[0] if l else None
Regarding idioms, there is an itertools recipe called nth.
From itertools recipes:
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)
If you want one-liners, consider installing a library that implements this recipe for you, e.g. more_itertools:
import more_itertools as mit
mit.nth([3, 2, 1], 0)
# 3
mit.nth([], 0) # default is `None`
# None
Another tool is available that only returns the first item, called more_itertools.first.
mit.first([3, 2, 1])
# 3
mit.first([], default=None)
# None
These itertools scale generically for any iterable, not only for lists.
for item in get_list():
return item
Frankly speaking, I do not think there is a better idiom: your is clear and terse - no need for anything "better". Maybe, but this is really a matter of taste, you could change if len(list) > 0: with if list: - an empty list will always evaluate to False.
On a related note, Python is not Perl (no pun intended!), you do not have to get the coolest code possible.
Actually, the worst code I have seen in Python, was also very cool :-) and completely unmaintainable.
By the way, most of the solution I have seen here do not take into consideration when list[0] evaluates to False (e.g. empty string, or zero) - in this case, they all return None and not the correct element.
my_list[0] if len(my_list) else None
Not sure how pythonic this is but until there is a first function in the library I include this in the source:
first = lambda l, default=None: next(iter(l or []), default)
It's just one line (conforms to black) and avoids dependencies.
Out of curiosity, I ran timings on two of the solutions. The solution which uses a return statement to prematurely end a for loop is slightly more costly on my machine with Python 2.5.1, I suspect this has to do with setting up the iterable.
import random
import timeit
def index_first_item(some_list):
if some_list:
return some_list[0]
def return_first_item(some_list):
for item in some_list:
return item
empty_lists = []
for i in range(10000):
empty_lists.append([])
assert empty_lists[0] is not empty_lists[1]
full_lists = []
for i in range(10000):
full_lists.append(list([random.random() for i in range(10)]))
mixed_lists = empty_lists[:50000] + full_lists[:50000]
random.shuffle(mixed_lists)
if __name__ == '__main__':
ENV = 'import firstitem'
test_data = ('empty_lists', 'full_lists', 'mixed_lists')
funcs = ('index_first_item', 'return_first_item')
for data in test_data:
print "%s:" % data
for func in funcs:
t = timeit.Timer('firstitem.%s(firstitem.%s)' % (
func, data), ENV)
times = t.repeat()
avg_time = sum(times) / len(times)
print " %s:" % func
for time in times:
print " %f seconds" % time
print " %f seconds avg." % avg_time
These are the timings I got:
empty_lists:
index_first_item:
0.748353 seconds
0.741086 seconds
0.741191 seconds
0.743543 seconds avg.
return_first_item:
0.785511 seconds
0.822178 seconds
0.782846 seconds
0.796845 seconds avg.
full_lists:
index_first_item:
0.762618 seconds
0.788040 seconds
0.786849 seconds
0.779169 seconds avg.
return_first_item:
0.802735 seconds
0.878706 seconds
0.808781 seconds
0.830074 seconds avg.
mixed_lists:
index_first_item:
0.791129 seconds
0.743526 seconds
0.744441 seconds
0.759699 seconds avg.
return_first_item:
0.784801 seconds
0.785146 seconds
0.840193 seconds
0.803380 seconds avg.
try:
return a[0]
except IndexError:
return None
def head(iterable):
try:
return iter(iterable).next()
except StopIteration:
return None
print head(xrange(42, 1000) # 42
print head([]) # None
BTW: I'd rework your general program flow into something like this:
lists = [
["first", "list"],
["second", "list"],
["third", "list"]
]
def do_something(element):
if not element:
return
else:
# do something
pass
for li in lists:
do_something(head(li))
(Avoiding repetition whenever possible)
Borrowing more_itertools.first_true code yields something decently readable:
def first_true(iterable, default=None, pred=None):
return next(filter(pred, iterable), default)
def get_first_non_default(items_list, default=None):
return first_true(items_list, default, pred=lambda x: x!=default)
Following code covers several scenarios by using lambda:
l1 = [1,2,3]
l2 = []
l3 = None
first_elem = lambda x: x[0] if x else None
print(first_elem(l1))
print(first_elem(l2))
print(first_elem(l3))
Using the and-or trick:
a = get_list()
return a and a[0] or None
Probably not the fastest solution, but nobody mentioned this option:
dict(enumerate(get_list())).get(0)
if get_list() can return None you can use:
dict(enumerate(get_list() or [])).get(0)
Advantages:
-one line
-you just call get_list() once
-easy to understand
My use case was only to set the value of a local variable.
Personally I found the try and except style cleaner to read
items = [10, 20]
try: first_item = items[0]
except IndexError: first_item = None
print first_item
than slicing a list.
items = [10, 20]
first_item = (items[:1] or [None, ])[0]
print first_item
How about this:
(my_list and my_list[0]) or None
Note: This should work fine for lists of objects but it might return incorrect answer in case of number or string list per the comments below.
You could use Extract Method. In other words extract that code into a method which you'd then call.
I wouldn't try to compress it much more, the one liners seem harder to read than the verbose version. And if you use Extract Method, it's a one liner ;)
Several people have suggested doing something like this:
list = get_list()
return list and list[0] or None
That works in many cases, but it will only work if list[0] is not equal to 0, False, or an empty string. If list[0] is 0, False, or an empty string, the method will incorrectly return None.
I've created this bug in my own code one too many times !
isn't the idiomatic python equivalent to C-style ternary operators
cond and true_expr or false_expr
ie.
list = get_list()
return list and list[0] or None
if mylist != []:
print(mylist[0])
else:
print(None)