我怎样才能一次浏览 100 个元素的列表？

Question

我得到了一个包含 958 个元素的列表。

myList = [1, 2, 3, 4, ..., 958]

我想取前 100 个元素，然后取下 100 个（从 100 到 200）等等。

我试过的：

sum = 0
ct = 0
for i in range(len(myList):
   sum = sum + myList[i]
   ct = ct + 1
   if ct == 100:
      ct = 0 
      print(sum)
      sum = 0

它在第 900 个元素之前一直有效。然后它不能对 myList 的最后 58 个元素求和，因为 ct 不会得到 100。

有任何想法吗？

Answer 1

一次遍历列表 100 并对切片求和。超出列表末尾的切片被正确处理：

myList = list(range(1,959))

for i in range(0,len(myList),100):
    sub = myList[i:i+100]
    print(f'sum of {sub[0]}..{sub[-1]} is {sum(sub)}')

输出：

sum of 1..100 is 5050
sum of 101..200 is 15050
sum of 201..300 is 25050
sum of 301..400 is 35050
sum of 401..500 is 45050
sum of 501..600 is 55050
sum of 601..700 is 65050
sum of 701..800 is 75050
sum of 801..900 is 85050
sum of 901..958 is 53911

Answer 2

因为只有在达到 100 的倍数时才打印总和，所以永远不会打印最后一组元素的总和，因为它不是 100 个元素。要获得最后一组，请在循环结束后打印总和。

您还可以使用 enumerate() 直接在循环中获取索引，而不是递增您自己的 ct 变量。

total = 0
for i, item in enumerate(myList, 1):
    total += item
    if i % 100 == 0:
        print(total)
        total = 0
# Print the final group
if i % 100 != 0:
    print(total)

我还将sum变量重命名为total，因为sum是内置函数的名称。

Answer 3

来自 itertools 模块的文档：

from itertools import zip_longest


def grouper(iterable, n, *, incomplete='fill', fillvalue=None):
    "Collect data into non-overlapping fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, fillvalue='x') --> ABC DEF Gxx
    # grouper('ABCDEFG', 3, incomplete='strict') --> ABC DEF ValueError
    # grouper('ABCDEFG', 3, incomplete='ignore') --> ABC DEF
    args = [iter(iterable)] * n
    if incomplete == 'fill':
        return zip_longest(*args, fillvalue=fillvalue)
    if incomplete == 'strict':
        return zip(*args, strict=True)
    if incomplete == 'ignore':
        return zip(*args)
    else:
        raise ValueError('Expected fill, strict, or ignore')


for century in grouper(myList, 100, fillvalue=0):
    print(sum(century))

Answer 4

您可以按如下方式对列表进行切片：

for k in range(0, len(mylist), 100):
  print(sum(mylist[k:k+100]))

这将打印 100 组元素的总和。如果列表长度不是 100 的精确倍数，最后的总和将是剩余元素的总和

Answer 5

尝试以下操作：

index = 1
slicesize = 100

start = 0
end = 100

myList = range(1, 958)

while True:
    myslice = myList[start:end]
    
    index += 1
    start = end
    end = index * slicesize
    if start > len(myList):
        break

Answer 6

我相信这应该可以解决问题。

import random

# generate list with N elements, each of which is a random integer in [0, 99]
N = 57
random.seed()
myList = [random.randint(0, 99) for i in range(0, N)]

# Calculate total number of prints before there will be less than num_elem
# elements remaining in the list.
num_elem = 10
total_prints = len(myList) // num_elem

for i in range(0, total_prints):
    index = i * num_elem
    print(sum(myList[index:index + num_elem]))

# if there are remaining elements, print their sum
if (len(myList) % num_elem) != 0:
    print(sum(myList[index + num_elem:]))

Answer 7

def n_elements_per_call(iterable, n=100):
    buffer = []
    for i in iterable:
        buffer.append(i)
        if len(buffer) < n:
            continue
        yield buffer
        buffer.clear()
    if buffer:
        yield buffer

my_list = list(range(959))
for nums in n_elements_per_call(my_list):
    print(nums)