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[英]Take the average and go through the list again. Count the amount of numbers greater than the average and the amount of numbers less than the average
[英]how to use while loop to determine the amount of numbers in a list that are greater the average
嗨,我需要以两种方式执行此任务:一种是使用for循环,另一种是使用while循环,但我没有剖析……。我编写的代码是:
A = [5,8,9,1,2,4]
AV = sum(A) / float(len(A))
count = 0
for i in A :
if i > AV :
count = count + 1
print ("The number of elements bigger than the average is: " + str(count))
count = 0
while float in A > AV:
count += 1
print ("The number of elements bigger than the average is: " + str(count))
您的代码确实未格式化。 通常:
for x in some_list:
... # Do stuff
等效于:
i = 0
while i < len(some_list):
... # Do stuff with some_list[i]
i += 1
问题是您在代码while float in A > AV:
中使用while float in A > AV:
使用。 在条件成立之前,while将起作用。 因此,一旦列表中遇到一些小于平均数的数字,循环就会退出。 因此,您的代码应为:
A = [5,8,9,1,2,4]
AV = sum(A) / float(len(A))
count = 0
for i in A : if i > AV :
count = count + 1
print ("The number of elements bigger than the average is: " + str(count))
count = 0
i = 0
while i < len(A):
if A[i] > AV: count += 1
i += 1
print ("The number of elements bigger than the average is: " + str(count))
我希望它能对您有所帮助:)并且我相信您知道为什么我要添加另一个变量i
。
A = [5,8,9,1,2,4]
AV = sum(A) / float(len(A))
count = 0
for i in A:
if i > AV:
count = count + 1
print ("The number of elements bigger than the average is: " + str(count))
count = 0
i = 0
while i < len(A):
if A[i] > AV:
count += 1
i += 1
print ("The number of elements bigger than the average is: " + str(count))
您可以使用类似下面的代码。 我对每个部分都进行了评论,解释了重要的部分。 请注意, while float in A > AV
中的while float in A > AV
在python中无效。 对于您的情况,应该通过索引或使用带有in
关键字的for
循环来访问列表的元素。
# In python, it is common to use lowercase variable names
# unless you are using global variables
a = [5, 8, 4, 1, 2, 9]
avg = sum(a)/len(a)
print(avg)
gt_avg = sum([1 for i in a if i > avg])
print(gt_avg)
# Start at index 0 and set initial number of elements > average to 0
idx = 0
count = 0
# Go through each element in the list and if
# the value is greater than the average, add 1 to the count
while idx < len(a):
if a[idx] > avg:
count += 1
idx += 1
print(count)
上面的代码将输出以下内容:
4.833333333333333
3
3
注意:我提供了列表理解的替代方法。 您也可以使用这段代码。
gt_avg = 0
for val in a:
if val > avg:
gt_avg += 1
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