3.4 fixed number of cycles II

We Continue to learn about the fixed number of loop statements .

Last time we talked about using loops to simplify input . Next , How to simplify the output ？

In a glance , It seems difficult to write it as flexible as input . The reason lies in input Only one data is entered for each execution , Here, a summary data is output together .

So if you want to write with a loop , First of all, we should split this statement ：

here , Output only output once , But the output value has been calculated . It's time to define variables again , We define a sum Save the sum .

The reason why it is split into this shape , Or is it for circular expression . Let's focus on this sum Sum up . We can imagine , Addition can only be done two by two , The specific accumulation calculation process is actually formed by the continuous accumulation of two numerical values . For example, first count two ：

You may see the law of circulation , But the first one is really different .

Let's use our brains , Write it in the same way as the following ：

The key here is to piece together a recyclable structure , Form a method that can be expressed circularly . therefore , We wrote the final code ：

sum = 0
for i in range(len(num)):
    sum = sum + num[i]
print(sum)

Students who still have questions can write down each execution process of this cycle by themselves , You will find that it is the corresponding multiple sum Cumulative statement ：

For lists with fixed values , We can also go straight through for Loop to read through each element , You do not need to access... By serial number ：

num = [1, 2, 3, 4, 5]
for i in num:
    print(i)

Output is ：
1
2
3
4
5

So the accumulation code just now can also be written as ：

sum = 0
for i in num:
    sum = sum + i
print(sum)

The output is the value of five list elements . here i No longer serial number , It is the corresponding value . The second line here for Statement means to fetch each list element in sequence , Save to i in , So you can for Pass in cycle i Get the value of the current list element .

The complete code is ：

num = [0] * 5
for i in range(len(num)):
    num[i] = int(input())

sum = 0
for i in range(len(num)):
    sum = sum + num[i]
print(sum)

perhaps ：

num = [0] * 5
for i in range(len(num)):
    num[i] = int(input())

sum = 0
for i in num:
    sum = sum + i
print(sum)

for Statements can also be used simply and flexibly , Such as ：

print([i for i in range(5)])

Output is ：[0, 1, 2, 3, 4], It can be understood as a continuous i Finally combined into a list .

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Just practiced print Output sum statement and continuous accumulation sum Of for Circulation is closely related , If you want to prevent subsequent code changes, you may for and print It changed accidentally sum, You can think about print and for Integrate together ：

num = [0] * 5
for i in range(len(num)):
    num[i] = int(input())

sum = 0
for i in range(len(num)):
    sum = sum + num[i]
else:
    print(sum)

there else Is with the for It's a whole , Throughout for At the end of the cycle , Automatically else The statement in , Therefore, the statements here usually express the closing function that needs to be performed at the end of the loop .

---

Next , We might as well adjust the code , See if you can continue to optimize ：

First We put all the variable definitions in front of us , This does not change the logical order of the code , So it's still true .

num = [0] * 5
sum = 0
for i in range(len(num)):
    num[i] = int(input())
for i in range(len(num)):
    sum = sum + num[i]
print(sum)

however , Does the existing two loops look like they can be merged ？

num = [0] * 5
sum = 0
for i in range(len(num)):
    num[i] = int(input())
    sum = sum + num[i]
print(sum)

Absolutely. , And the function and effect are the same .

however , We observe carefully again , Do you find that there is a waste of steps ？ In circulation , Each time we put the entered value into the list element , Then add the values of the list elements to sum, So why not just add the input values to sum What about China? ？

num = [0] * 5
sum = 0
for i in range(len(num)):
    sum = sum + int(input())
print(sum)

in fact , Absolutely. . The code is written here , We can even do the same without lists at all ：

sum = 0
for i in range(5):
    sum = sum + int(input())
print(sum)

Of course , This does not mean that using lists makes no sense . Although the same effect can be achieved at this time , But all kinds of original input data have been lost , in other words , If you need to retrieve these entered values again later , The use of lists can still provide continuous access to each input value . therefore , We also need to flexibly choose the implementation of the code according to our own needs .

Supporting learning resources 、 MOOC video ：

Python Big data analysis - Shu Qing Li https://www.njcie.com/python/