Could not find a version that satisfies the requirement

| | |

I"m installing several Python packages in Ubuntu 12.04 using the following requirements.txt file:

numpy>=1.8.2,<2.0.0
matplotlib>=1.3.1,<2.0.0
scipy>=0.14.0,<1.0.0
astroML>=0.2,<1.0
scikit-learn>=0.14.1,<1.0.0
rpy2>=2.4.3,<3.0.0

and these two commands:

$ pip install --download=/tmp -r requirements.txt
$ pip install --user --no-index --find-links=/tmp -r requirements.txt

(the first one downloads the packages and the second one installs them).

The process is frequently stopped with the error:

  Could not find a version that satisfies the requirement <package> (from matplotlib<2.0.0,>=1.3.1->-r requirements.txt (line 2)) (from versions: )
No matching distribution found for <package> (from matplotlib<2.0.0,>=1.3.1->-r requirements.txt (line 2))

which I fix manually with:

pip install --user <package>

and then run the second pip install command again.

But that only works for that particular package. When I run the second pip install command again, the process is stopped now complaining about another required package and I need to repeat the process again, ie: install the new required package manually (with the command above) and then run the second pip install command.

So far I"ve had to manually install six, pytz, nose, and now it"s complaining about needing mock.

Is there a way to tell pip to automatically install all needed dependencies so I don"t have to do it manually one by one?

Add: This only happens in Ubuntu 12.04 BTW. In Ubuntu 14.04 the pip install commands applied on the requirements.txt file work without issues.

Could not find a version that satisfies the requirement find: Questions

Finding the index of an item in a list

5 answers

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

3740

Answer #1

>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514
 

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

3740

Answer #2

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
 ...

 |
 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value
 |

which will often lead you to the method you are looking for.

3740

Answer #3

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":
        print(i)

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop

Could not find a version that satisfies the requirement repeat: Questions

Create list of single item repeated N times

5 answers

I want to create a series of lists, all of varying lengths. Each list will contain the same element e, repeated n times (where n = length of the list).

How do I create the lists, without using a list comprehension [e for number in xrange(n)] for each list?

618

Answer #1

You can also write:

[e] * n

You should note that if e is for example an empty list you get a list with n references to the same list, not n independent empty lists.

Performance testing

At first glance it seems that repeat is the fastest way to create a list with n identical elements:

>>> timeit.timeit("itertools.repeat(0, 10)", "import itertools", number = 1000000)
0.37095273281943264
>>> timeit.timeit("[0] * 10", "import itertools", number = 1000000)
0.5577236771712819

But wait - it"s not a fair test...

>>> itertools.repeat(0, 10)
repeat(0, 10)  # Not a list!!!

The function itertools.repeat doesn"t actually create the list, it just creates an object that can be used to create a list if you wish! Let"s try that again, but converting to a list:

>>> timeit.timeit("list(itertools.repeat(0, 10))", "import itertools", number = 1000000)
1.7508119747063233

So if you want a list, use [e] * n. If you want to generate the elements lazily, use repeat.

What is the best way to repeatedly execute a function every x seconds?

5 answers

DavidM By DavidM

I want to repeatedly execute a function in Python every 60 seconds forever (just like an NSTimer in Objective C). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.

In this question about a cron implemented in Python, the solution appears to effectively just sleep() for x seconds. I don"t need such advanced functionality so perhaps something like this would work

while True:
    # Code executed here
    time.sleep(60)

Are there any foreseeable problems with this code?

391

Answer #1

If your program doesn"t have a event loop already, use the sched module, which implements a general purpose event scheduler.

import sched, time
s = sched.scheduler(time.time, time.sleep)
def do_something(sc): 
    print("Doing stuff...")
    # do your stuff
    s.enter(60, 1, do_something, (sc,))

s.enter(60, 1, do_something, (s,))
s.run()

If you"re already using an event loop library like asyncio, trio, tkinter, PyQt5, gobject, kivy, and many others - just schedule the task using your existing event loop library"s methods, instead.

391

Answer #2

Lock your time loop to the system clock like this:

import time
starttime = time.time()
while True:
    print "tick"
    time.sleep(60.0 - ((time.time() - starttime) % 60.0))

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