Question

我想使用for循环来找到python2.7

的每个三元素的最小差异

这是我的test.csv数据

E1-2    u7  4
E1-2    u7  7
E1-2    u7  8
F(1)-1  u7  3
F(1)-1  u7  9
F(1)-1  u7  8
..     ..  ..

我希望得到以下结果

E1-2    u7  7
E1-2    u7  8
F(1)-1  u7  8
F(1)-1  u7  9
..     ..  ..

因为在组“E1-2”中，最小差异元素是7和8

8-4=5
8-7=1  # the minimal difference
7-4=3

这是我的感冒，但有一些问题。谁能帮我修改一下？感谢。

from itertools import tee, izip
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)`

def mins (datas):
mi=min(pairwise(sorted(datas)), key=lambda x: x[1] - x[0])
return list(mi)`

import pandas as pd
qrt=pd.read_csv("test.csv")

def cul(dat,sam):
#results=pd.DataFrame()
a=dat.iloc[0:2]
b=pd.Series(mins(dat[sam].tolist()))
a.loc[:,'D']=b
return a`

for i in range(0,len(qrt["Ct"])-1,3):
     results=pd.DataFrame()
 group=qrt[i:i+3]
 c=cul(group,"Ct")
 results=results.append(c)
 print results.head()

输出：

  Sample Detector      Ct       D
0   E1-2     u7  4  7
1   E1-2     u7  7  8
  Sample Detector      Ct   D
3  F(1)-1     u7  3 NaN
4  F(1)-1     u7  9 NaN

Answer 1

from collection import defaultdict
import itertools
import csv

d = defaultdict(list)
with open("test.csv", newline='') as f:
    r = csv.reader(f)
    for row in r:
         row[2] = float(row[2])
         d[row[0]].append(row)

for k, v in d.items():
    a, b = min(itertools.combinations(v, 2), key=lambda x: abs(x[0][2]-x[1][2]))
    print(a)
    print(b)

我们将使用itertools.combinations查找所有元素对，然后找到给出最小差异的对并打印该对中的每个元素。您可以将这些print替换为您想要对该数据执行的操作

Answer 2

data = [("a", 3), ("a", 5), ("a", 8), ("b", 18), ("b", 13), ("b", 19)]

indexes = set(index for index, _ in data)

index_values = {index:[value for i, value in data if i == index] for index in indexes}

from itertools import combinations

def difference(pair):
    x, y = pair
    return abs(x - y)

def smallest_difference(values):
    min_diff, min_pair = min((difference(pair), pair) for pair in combinations(values,2))
    return min_pair

index_min_pairs = {index:smallest_difference(values) for index, values in index_values.items()}

answer = [(index, min_value) for index in index_min_pairs for min_value in index_min_pairs[index]]

for循环使用python2.7找到每个三元素的最小差异

2 个答案: