How to install 2 Anacondas (Python 2 and 3) on Mac OS

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I"m relatively new in Mac OS. I"ve just installed XCode (for c++ compiler) and Anaconda with the latest Python 3 (for myself). Now I"m wondering how to install properly second Anaconda (for work) with Python 2?

I need both versions to work with iPython and Spyder IDE. Ideal way is to have totally separate Python environments. For example, I wish I could write like conda install scikit-learn for Python 3 environment and something like conda2 install scikit-learn for Python 2.

How to install 2 Anacondas (Python 2 and 3) on Mac OS sep: Questions

sep

How to print number with commas as thousands separators?

5 answers

I am trying to print an integer in Python 2.6.1 with commas as thousands separators. For example, I want to show the number 1234567 as 1,234,567. How would I go about doing this? I have seen many examples on Google, but I am looking for the simplest practical way.

It does not need to be locale-specific to decide between periods and commas. I would prefer something as simple as reasonably possible.

929

Answer #1

Locale unaware

"{:,}".format(value)  # For Python ‚â•2.7
f"{value:,}"  # For Python ‚â•3.6

Locale aware

import locale
locale.setlocale(locale.LC_ALL, "")  # Use "" for auto, or force e.g. to "en_US.UTF-8"

"{:n}".format(value)  # For Python ‚â•2.7
f"{value:n}"  # For Python ‚â•3.6

Reference

Per Format Specification Mini-Language,

The "," option signals the use of a comma for a thousands separator. For a locale aware separator, use the "n" integer presentation type instead.

929

Answer #2

I got this to work:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, "en_US")
"en_US"
>>> locale.format("%d", 1255000, grouping=True)
"1,255,000"

Sure, you don"t need internationalization support, but it"s clear, concise, and uses a built-in library.

P.S. That "%d" is the usual %-style formatter. You can have only one formatter, but it can be whatever you need in terms of field width and precision settings.

P.P.S. If you can"t get locale to work, I"d suggest a modified version of Mark"s answer:

def intWithCommas(x):
    if type(x) not in [type(0), type(0L)]:
        raise TypeError("Parameter must be an integer.")
    if x < 0:
        return "-" + intWithCommas(-x)
    result = ""
    while x >= 1000:
        x, r = divmod(x, 1000)
        result = ",%03d%s" % (r, result)
    return "%d%s" % (x, result)

Recursion is useful for the negative case, but one recursion per comma seems a bit excessive to me.

sep

How would you make a comma-separated string from a list of strings?

5 answers

mweerden By mweerden

What would be your preferred way to concatenate strings from a sequence such that between every two consecutive pairs a comma is added. That is, how do you map, for instance, ["a", "b", "c"] to "a,b,c"? (The cases ["s"] and [] should be mapped to "s" and "", respectively.)

I usually end up using something like "".join(map(lambda x: x+",",l))[:-1], but also feeling somewhat unsatisfied.

612

Answer #1

my_list = ["a", "b", "c", "d"]
my_string = ",".join(my_list)
"a,b,c,d"

This won"t work if the list contains integers


And if the list contains non-string types (such as integers, floats, bools, None) then do:

my_string = ",".join(map(str, my_list)) 

How to install 2 Anacondas (Python 2 and 3) on Mac OS StackOverflow: Questions

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