1

我希望在 python 中创建一个程序，要求用户以 [] 格式输入数字列表。

然后它必须计算并显示列表中有多少这些数字介于 1 和 10、10 和 20、20 和 30 等之间。

这是我到目前为止所拥有的。

my_num = [int(i) for i in raw_input("Enter numbers... ").split(",")]

the_list = eval(my_num)

number = 0

if 1<= my_num and my_num <=10:
    number = number + 1
4

6 回答 6

5 
import itertools

numbers = sorted(map(int, raw_input().split(",")))
for k, g in itertools.groupby(numbers, lambda x: x // 10):
    print k, list(g)

这将为您提供一组数字，您可以通过调用来计算组中的实例len()。例如，替换

    # print k, list(g)
    print k, len(list(g))

重要的是首先对数字进行排序。

以此作为输入：

1, 5, 6, 89, 43, 43, 25, 76, 12, 32, 23, 25, 27, 13, 5, 7

0 [1, 5, 5, 6, 7]
1 [12, 13]
2 [23, 25, 25, 27]
3 [32]
4 [43, 43]
7 [76]
8 [89]

或者使用 len 调用的输出：

0 5
1 2
2 4
3 1
4 2
7 1
8 1

然后可以像这样格式化：

print "%d - %d: %d" % (k * 10 + 1, (k + 1) * 10, len(list(g))

产量：

 1 - 10: 5
11 - 20: 2
21 - 30: 4
31 - 40: 1
41 - 50: 2
71 - 80: 1
81 - 90: 1
于 2012-08-30T22:48:40.330 回答
2

使用 aCounter是有效的，因为您只需要遍历列表一次

>>> L = [1, 5, 6, 89, 43, 43, 25, 76, 12, 32, 23, 25, 27, 13, 5, 7]
>>> from collections import Counter
>>> Counter(x//10 for x in L)
Counter({0: 5, 2: 4, 1: 2, 4: 2, 3: 1, 7: 1, 8: 1})
>>> sorted(Counter(x//10 for x in L).items())
[(0, 5), (1, 2), (2, 4), (3, 1), (4, 2), (7, 1), (8, 1)]

>>> for k,v in sorted(Counter(x//10 for x in L).items()):
...     print "%d - %d: %d"%(k*10, k*10+9, v)
...
0 - 9: 5
10 - 19: 2
20 - 29: 4
30 - 39: 1
40 - 49: 2
70 - 79: 1
80 - 89: 1
于 2012-08-30T23:15:35.163 回答
1

使用len()和列表理解：

>>> lis=[1, 5, 6, 89, 43, 43, 25, 76, 12, 32, 23, 25, 27, 13, 5, 7]

>>> [(i,[x for x in lis if x//10==i]) for i in range(0,(max(lis)//10)+1)]
[(0, [1, 5, 6, 5, 7]), (1, [12, 13]), (2, [25, 23, 25, 27]), (3, [32]), (4, [43, 43]), (5, []), (6, []), (7, [76]), (8, [89])]

长度使用len()：

>>> [(i,len([x for x in lis if x//10==i])) for i in range(0,(max(lis)//10)+1)]
[(0, 5), (1, 2), (2, 4), (3, 1), (4, 2), (5, 0), (6, 0), (7, 1), (8, 1)]
于 2012-08-30T22:45:41.120 回答
0

如果我理解，您想确定不同范围内的数字数量：[1;10]​​、[11; 20]；[21; 30], ..., [91; 100]

# Example of user input
your_list = [2, 10, 93, 1, 20]

# All the ranges' right part (sorry if my english is bad). my_ranges = [10, 20, 30, ..., 100]
my_ranges = []

# Array determining how many of those numbers in the list are in [1; 10], [11; 20], ..., [91; 100]
numbers = []

for i in range(1, 11):
    my_ranges.append(i*10)
    numbers.append(0)

for i in range(0, len(your_list)):
    out = False
    j = 0
    while not out and j<len(my_ranges):
        if your_list[i] <= my_ranges[j]:
            out = True
            numbers[j]+=1
        j+=1

for i in range(0, len(numbers)):
    print "Number of numbers in ["+str(my_ranges[i]+1-10)+";"+str(my_ranges[i])+"] = "+str(numbers[i])
于 2012-08-30T23:07:28.557 回答
0

不一定是最简单的方法，但ast.literal_eval可以是eval.

至于计算每组十个数字中有多少个，该itertools模块有一个itertools.groupby

import ast
import itertools


the_input = raw_input("Enter numbers... ")

# Lazy way to parse '[1,2,3]' into [1,2,3]
the_list = ast.literal_eval(the_input)

# Or, parse it manually (this is probably better in this case):
the_list = [int(x) for x in the_input.strip("[]").split(",")]

# Group by dividing number by 10, the result as an integer (15//10==1)
# This means 0-9 are in "group 0", 10-19 are in "group 1" etc
groups = itertools.groupby(the_list, key = lambda k: k // 10)

# Show stats for each block of ten
for group_count, group_members in groups:
    print "There were {count} numbers in {start}-{end}".format(
        count = len(list(group_members)),
        start = group_count * 10, # 15/10 becomes 1, then we do 1*10 for 10
        end = group_count * 10 + 9) # and 1*10+9 to get 19 (displayed as "in 10-19"

示例运行：

$ python script.py
Enter numbers... [1,2, 11,12,13, 22,23, 44,45,46,47]
There were 2 numbers in 0-9
There were 3 numbers in 10-19
There were 2 numbers in 20-29
There were 4 numbers in 40-49

要将其调整为组 1-10、11-20 等，只需group =像这样更改行线：

groups = itertools.groupby(the_list, key = lambda k: k // 10 + 1)

或者，更简单一点：

the_input = raw_input("Enter numbers... ")

# Parse "[1,2,3]" into a list of numbers
the_list = [int(x) for x in the_input.strip("[]").split(",")]

group_counter = {}

for number in the_list:
    group_start = (number // 10) * 10
    group_end = group_start + 9
    group_name = "%s-%s" % (group_start, group_end)

    group_counter.setdefault(group_name, 0)
    group_counter[group_name] += 1


for name, count in group_counter.items():
    # Note that the sorting is wrong
    print "There were %s in %s" % (count, name)

collections.Counter如果您使用 Python 2.7 或更高版本，也可以使用

于 2012-08-30T23:11:55.217 回答
0
  1. 计算列表中的数字总数
  2. 列表中所有数字的总和和平均值
  3. 计算列表中所有奇数的总和
  4. 计算列表中所有偶数的总和
  5. 在列表中找到最大的数字
  6. 在列表中找到最小的数字

显示具有适当标题的所有值。

listNo = [6,8,10,44,33,21,7,1,0,2]
c = 0
s = 0
avg = 0
sOdd = 0
sEven = 0
cOdd = 0
cEven = 0
for i in listNo :
    c += 1
    s = s+i
    avg = s/c
    if i % 2 == 0 :
        sEven = sEven + i
        cEven = cEven + 1
    else :
        sOdd = sOdd + i
        cOdd = cOdd + 1
print ("total number of numbers in the list  : ", c)
print("sum of all numbers : ",s)
print("average of all numbers : ",avg)
print("count odd numbers : ",cOdd)
print("sum of odd numbers : ",sOdd)
print("count even numbers : ",cEven)
print("sum of odd numbers : ",sEven)
print("largest number in the list : " ,max(listNo))
print("smallest number in the list  : ",min(listNo))
于 2021-04-14T06:04:33.270 回答