字典通过更新合并，但如果值存在则不覆盖

本文介绍了字典通过更新合并，但如果值存在则不覆盖的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

If I have 2 dicts as follows:

d1 = {'a': 2, 'b': 4} d2 = {'a': 2, 'b': ''}

In order to 'merge' them:

dict(d1.items() + d2.items())

results in

{'a': 2, 'b': ''}

But what should I do if I would like to compare each value of the two dictionaries and only update d2 into d1 if values in d1 are empty/None/''?

When the same key exists, I would like to only maintain the numerical value (either from d1 or d2) instead of the empty value. If both values are empty, then no problems maintaining the empty value. If both have values, then d1-value should stay.

i.e.

d1 = {'a': 2, 'b': 8, 'c': ''} d2 = {'a': 2, 'b': '', 'c': ''}

should result in

{'a': 2, 'b': 8, 'c': ''}

where 8 is not overwritten by ''.

解决方案

Just switch the order:

z = dict(d2.items() + d1.items())

By the way, you may also be interested in the potentially faster update method.

In Python 3, you have to cast the view objects to lists first:

z = dict(list(d2.items()) + list(d1.items()))

If you want to special-case empty strings, you can do the following:

def mergeDictsOverwriteEmpty(d1, d2): res = d2.copy() for k,v in d2.items(): if k not in d1 or d1[k] == '': res[k] = v return res