Python: assigning variable name versus variable value to dictionary value

Question

Can someone explain why these two different code samples give different outputs? I'm confused on why the first sample appends "sunny" and "cloud" to all months, which is not what I want. I want the output of the second sample. However, I want to assign the variable "forecast", not its value to weatherReport["months"][month]. Is there a way to get the output of Sample 2 this way?

Sample 1:

weatherReport = {"months": {}}


forecast = []

months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]

for month in months[1:]:
    weatherReport["months"][month] = forecast

weatherReport["months"]["June"].append("sunny")
weatherReport["months"]["December"].append("cloudy")

print weatherReport

Output for Sample 1:

{'months': {'February': ['sunny', 'cloudy'], 'October': ['sunny', 'cloudy'], 'March': ['sunny', 'cloudy'], 'August': ['sunny', 'cloudy'], 'May': ['sunny', 'cloudy'], 'December': ['sunny', 'cloudy'], 'June': ['sunny', 'cloudy'], 'September': ['sunny', 'cloudy'], 'April': ['sunny', 'cloudy'], 'July': ['sunny', 'cloudy'], 'November': ['sunny', 'cloudy']}}

Sample 2:

weatherReport = {"months": {}}


forecast = []

months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]

for month in months[1:]:
    weatherReport["months"][month] = []


weatherReport["months"]["June"].append("sunny")
weatherReport["months"]["December"].append("cloudy")

print weatherReport

Output for Sample 2:

{'months': {'February': [], 'October': [], 'March': [], 'August': [], 'May': [], 'December': ['cloudy'], 'June': ['sunny'], 'September': [], 'April': [], 'July': [], 'November': []}}

score 2 · Accepted Answer · answered May 22 '16 at 08:40

2

In second code, you assign a brand new independant list for each month. So when you later append values to those lists, you append only to the proper list.

In first code, you assign the same variable to each month. In Python it means that each month has a reference to same list. So when you add a value to any of those list, in reality you add it to the forecast list, to which each month points. So the result.

Under the hood the cause is that a list is mutable, that means that you can modify an object which other objects point to. You will not be able to reproduce that with non mutable objects like strings.

answered May 22 '16 at 08:40

Serge Ballesta

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I see. Is there any other way of getting the same solution as the 2nd code, or is that the only way? – user3226932 May 22 '16 at 08:46
@user3226932 List copying: forecast[:] and you don't have a reference but a copy. – palsch May 22 '16 at 09:05

score 1 · Answer 2 · answered May 22 '16 at 08:49

In sample 1 you are assigning the same reference to all values in the dict; so, when you change the value of that reference, it is reflected across all values in the dictionary. In the second sample, you have a unique list for each dictionary instance.

Python: assigning variable name versus variable value to dictionary value

2 Answers2