如何允许用户脱离字典理解能力？

Question

I was playing with a simple dictionary comprehension code, and have altered it to allow for automatically generated keys (sequential numbers, which can be swapped out by way of code if desired) and user input for the values. I also have it to where the user can input the keys and the values.

My problem is figuring out how to allow the user to break out of it; ie, if they choose 100 for the number of entries, but find they only need 27, there needs to be a way to enter a command to end the process.

The third example is where I tried putting a condition on the end, but failed to figure it out. I knew it wouldn't work when I did it - it allows the user to go through one run.

"""
Example 1:
Auto-generated sequential numerical keys with user input for values:
"""

variable1=int(input("please select number of entries: ")) 

d1 = {x : input("Provide data: ") for x in range(1,variable1 + 1)}

"""
Example 2:
User input for keys and values:
"""
variable1=int(input("please select number of entries: ")) 

d1 = {input("Please provide key: "): input("Please provide data: " for x in range(1,variable1 + 1)}


"""
Example 3:
This shows where I tried to put the condition, which failed:
"""
variable1=int(input("please select number of entries: ")) 

d1 = {x : input("Provide data: ") for x in range(1,variable1 + 1) if x != 0}

Answer 1

You should not be using a dictionary comprehension here. Could you? Sure. Is it going to help you learn stuff as a new user to Python? No.

Take for example chepner's answer. It probably works. I'm not going to test it. Is it readable and easy to understand? Maybe. If your goal is to get a job, a lot of people look for more than just a right answer. There are many, many, many ways to get a right answer. If your answer is hard to comprehend for an interviewer, I would definitely think about writing the answer differently. I would recommend taking a look at sub-reddits for interview/programming help as well. You may also want to take a look at cracking the coding interview for preparing for technical interviews since they are very different from a normal interview.

In terms of your actual question:

    In [54]: d1 = {}
    ...: for i in range(variable1):
    ...:     data = input('provide data:')
    ...:     if data =='x':
    ...:         break
    ...:     else:
    ...:         data = int(data)
    ...:     d1[i+1] = data
    ...:
    ...:
provide data:12
provide data:13
provide data:14
provide data:x

In [55]: d1
Out[55]: {1: 12, 2: 13, 3: 14}

This will cycle up to variable1 , but if the user provides x , the loop will break. Just because something can be done in one line doesn't mean it should ;)

Answer 2

The 2-argument form of iter will call a function until it returns a particular variable. Specifically, iter(input, "0") will produce a stream of non-zero integer strings, which you can zip with your range:

d1 = {x : y for x, y in zip(
                            range(1,100),
                            iter(lambda: input("Provide data "), "0")
                        )
}

Or, use enumerate to generate an infinite stream of numbered calls to input which you can filter with islice :

from itertools import islice


d1 = {x: y for x, y in islice(
                           enumerate(
                               iter(lambda: input("Provide data "), "0"),
                               1
                           ),
                           100
                       )
}

In either case, because you have an unfiltered stream of tuples, you can pass that directly to dict instead of using a comprehension.

d1 = dict(zip(range(1,100), iter(lambda: input("Provide data "), "0")))
d1 = dict(islice(enumerate(iter(lambda: input("Provide data "), "0"), 1), 100))

---

For those that (justifiably) dislike the Lisp-like nature of all the nested function calls, you can easily break this into several lines for readability. For example,

responses = iter(lambda: input("Provide data "), "0")
d1a = dict(zip(range(1,100), responses))
d1b = dict(islice(enumerate(responses, 1)), 100)   

Or, if you prefer a traditional loop:

d = {}
for k in range(1, 100):  # or itertools.count(1) for unbounded input
    v = input("Provide data ")
    if v == "0":
        break
    d[k] = v