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一种更简单的方式来引用所需字典(Python)的索引?

[英]An easier way of referring to the index of a desired dictionary (Python)?

Question feels like it's phrased poorly, feel free to adjust it if you agree and know how better to phrase it. 问题感觉它的措辞很差,如果你同意的话可以随意调整它,并且知道如何更好地表达它。

I have the following code: 我有以下代码:

def owned_calendars(cal_items):
    """Returns only the calendars in which the user is marked as "owner"

    """
    owner_cals = []
    for entry in cal_items:
        if entry['accessRole'] == "owner":
            owner_cals.append(cal_items[cal_items.index(entry)])

    return owner_cals

cal_items is a list of dictionaries cal_items是一个dictionaries list

In the line where I have written owner_cals.append(cal_items[cal_items.index(entry)]) I'm trying to append the dictionaries that have the property accessRole = owner . 在我编写owner_cals.append(cal_items[cal_items.index(entry)])我正在尝试附加具有属性accessRole = owner的字典。

The line just seems super long and clunky, and I'm wondering if there's an easier/more intuitive way to do it? 这条线似乎超长而且笨重,我想知道是否有更容易/更直观的方法来做到这一点?

Try this. 尝试这个。 You can do this in one line using list comprehension . 您可以使用列表推导在一行中执行此操作。

owner_cals = [x for x in cal_items if x["access_role"]=="owner"]

You can also use enumerate method. 您还可以使用enumerate方法。

owner_cals = [j for i,j in enumerate(cal_items) if j["access_role"]=="owner"]

Also, remember .index() returns the lowest index where item is found . 另外,请记住.index()返回找到item的最低索引

["foo", "bar", "baz", "bar"].index("bar") will always return 1. ["foo", "bar", "baz", "bar"].index("bar")将始终返回1。

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