给出以下Python列表:
l1 = [0,1000,5000,10000,20000,30000,40000,50000] #8 values, 7 intervals
v1 = [1,2,3,4,5,6,7] #7 values
v2 = [a,b,c,d,e,f,g] #7 letters
我想检查数字是否包含在任何间隔中,并根据该间隔返回另一个值。
示例:
1. My test value is 1111
2. It belongs to the second interval: 1000 < 1111 < 5000
3. Hence I need to return b
我将通过以下方式解决这个问题:
l1
的块我可以通过查看每对连续的数字来创建它的块:
def chunker(seq, size):
return (seq[pos:pos + size] for pos in range(0, len(seq)))
for group in chunker(l1, 2):
print(group)
此返回:
[0, 1000]
[1000, 5000]
[5000, 10000]
[10000, 20000]
[20000, 30000]
[30000, 40000]
[40000, 50000]
[50000]
我的问题:
您可以zip
列出l1
和l1[1:]
,它们将以您所需的方式进行间隔。对于一些额外的指针,由于可以对间隔进行排序,因此您可以将二进制搜索合并到其中,我将让您进行优化。该算法的当前运行时间为O(n)
。
from __future__ import print_function
l1 = [0,1000,5000,10000,20000,30000,40000,50000]
v2 = ['a','b','c','d','e','f','g']
element = 1111
interval = []
for left,right in zip(l1,l1[1:]):
if left <= element <= right:
interval = [left,right]
break
print("answer is" ,interval)
输出
answer is [1000, 5000]
仅适用于len 2的块。
l1 = [0,1000,5000,10000,20000,30000,40000,50000] #8 values, 7 intervals
# v1 = [1,2,3,4,5,6,7] #7 values
# v2 = [a,b,c,d,e,f,g] #7 letters
def chunker(seq, size):
return (seq[pos:pos + size] for pos in range(0, len(seq)))
def chunk_with_value(list_of_chunks, value):
"""returns the chunk if the value is inside the range"""
for chunk in list_of_chunks:
if chunk[0] < value < chunk[1]:
return chunk
chunks = chunker(l1, 2)
print(chunk_with_value(chunks, 1111))
# [1000, 5000]
直接从itertools食谱书中:
def get_thing(value):
def pairwise(iterable):
from itertools import tee
a, b = tee(iterable)
next(b, None)
return zip(a, b)
interval_ranges = [
0,
100,
500,
1000
]
# There are four interval ranges, so three intervals.
things = [
"A", # 0-100
"B", # 100-500
"C" # 500-1000
]
for (begin, end), thing in zip(pairwise(interval_ranges), things):
if begin <= value < end: # modify this to suit your needs. Is the range inclusive/exclusive?
return thing
return None
def main():
thing = get_thing(400)
print(thing)
return 0
if __name__ == "__main__":
import sys
sys.exit(main())
输出:
B
我相信您需要获取相应的间隔索引,并使用该索引来查询v2
。这应该做到:
l1 = [0,1000,5000,10000,20000,30000,40000,50000] #8 values, 7 intervals
v2 = ['a','b','c','d','e','f','g'] #7 letters
def intervals(l):
for i in range(len(l)-1):
yield i, l[i:i+2]
def interval_value(val, interval_list, value_list):
for i, interval in intervals(interval_list):
if interval[0] <= val <= interval[1]:
return value_list[i]
print(interval_value(1111, l1, v2))
print(interval_value(0, l1, v2))
print(interval_value(51000, l1, v2))
print(interval_value(40000, l1, v2))
输出:
b
a
None
f
您不需要v1
值-您可以直接使用v2
上的索引
尽管其他答案很有用,但几乎所有答案都发现从v2
输出的间隔不是期望的>
l1 = [0,1000,5000,10000,20000,30000,40000,50000]
v2 = ['a','b','c','d','e','f','g']
element = 1111
def get_interval(l1):
for index, left,right in zip(range(len(l1)), l1, l1[1:]):
if left <= element <= right:
return v2[index]
>>> print("answer is:" ,get_interval(element))
>>> answer is: b