python与make中的$(其中)相等

Question

将Makefile移植到dodo.py，我想在路径上找到"nikola“，而不是直接调用它

在Makefile中，我会做一些类似的事情

NIKOLA = $(which nikola)

但是，我似乎无法找到doit/dodo/python与我当前的文件类似的如下所示

import subprocess

Nikola = subprocess.Popen("nikola", stdout=subprocess.PIPE, shell=True)

def task_build():
    return {
        'actions': ['{0} build'.format(Nikola)],
    }

但是，运行task_build会返回以下内容

########################################
Execution aborted.
Task 'build': invalid 'actions' type. got:<subprocess.Popen object at 0x10111f810> <class 'subprocess.Popen'>
(env)

Answer 1

试试这个：

>>> import subprocess
>>> subprocess.check_output(['which','nikola'])
'/Users/elyase/miniconda/bin/nikola\n'

就你而言：

def task_build():
    return {
        'actions': ['{0} build'.format(Nikola)],
    }

>>> Nikola = subprocess.check_output(['which','nikola']).rstrip()
>>> task_build()
{'actions': ['/Users/elyase/miniconda/bin/nikola build']}