每行每個項目的迭代置換

-1

我想操縱數據使用ggnet進行網絡分析。每行每個項目的迭代置換

數據集採用CSV形式，看起來像這樣：

offers 
{9425, 5801, 18451, 17958, 16023, 7166} 
{20003, 17737, 4031, 5554} 
{19764, 5553, 5554}

我想破了陣，並反覆以排列的所有項目每行一對2。所以最終輸出的應該看起來像：

print list(itertools.permutations([1,2,3,4], 2)) per row to create: 

(9425, 5801) 
(9425, 18451) 
(9425, 17958) 
(9425, 16023) 
(9425, 7166) 
(5801, 18451) 
(5801, 17958) 
(5801, 16023) 
(5801, 7166) 
...

我可以使用R或Python來做到這一點。任何建議來解決這個問題？

來源

2017-08-21 tmhs

你已經找到了'itertools.permutations'。你還在找什麼？ –

這是一個關於導入數據和/或排列的問題嗎？ –

另一條R解決方案，假設有在文件中更多的行。

# read in csv file as list of integers (each row in csv = 1 list element) 
offers <- readLines("offers.csv") %>% strsplit(",") %>% lapply(as.integer) 

# create permutation pairs for each element in the list 
permutation.list <- lapply(seq_along(offers), function(i) {t(combn(offers[[i]], m = 2))}) 

# combine all permutation pairs into 1 data frame 
permutation.data.frame <- plyr::ldply(permutation.list, data.frame)

下面是根據所提供的樣本數據的結果：

> permutation.list 
[[1]] 
     [,1] [,2] 
[1,] 9425 5801 
[2,] 9425 18451 
[3,] 9425 17958 
[4,] 9425 16023 
[5,] 9425 7166 
[6,] 5801 18451 
[7,] 5801 17958 
[8,] 5801 16023 
[9,] 5801 7166 
[10,] 18451 17958 
[11,] 18451 16023 
[12,] 18451 7166 
[13,] 17958 16023 
[14,] 17958 7166 
[15,] 16023 7166 

[[2]] 
     [,1] [,2] 
[1,] 20003 17737 
[2,] 20003 4031 
[3,] 20003 5554 
[4,] 17737 4031 
[5,] 17737 5554 
[6,] 4031 5554 

[[3]] 
     [,1] [,2] 
[1,] 19764 5553 
[2,] 19764 5554 
[3,] 5553 5554 

> permutation.data.frame 
     X1 X2 
1 9425 5801 
2 9425 18451 
3 9425 17958 
4 9425 16023 
5 9425 7166 
6 5801 18451 
7 5801 17958 
8 5801 16023 
9 5801 7166 
10 18451 17958 
11 18451 16023 
12 18451 7166 
13 17958 16023 
14 17958 7166 
15 16023 7166 
16 20003 17737 
17 20003 4031 
18 20003 5554 
19 17737 4031 
20 17737 5554 
21 4031 5554 
22 19764 5553 
23 19764 5554 
24 5553 5554

來源

2017-08-21 06:32:07

您可以在R試試這個：

a <- c(9425, 5801, 18451, 17958, 16023, 7166) 
b <- c(20003, 17737, 4031, 5554) 
c <- c(19764, 5553, 5554) 

rbind(t(combn(a,2)), 
t(combn(b,2)), 
t(combn(c,2)))

來源

2017-08-21 06:08:08

t(do.call(cbind,mapply(combn,list(a,b,c),2))) 
     [,1] [,2] 
[1,] 9425 5801 
[2,] 9425 18451 
[3,] 9425 17958 
[4,] 9425 16023 
[5,] 9425 7166 
[6,] 5801 18451 
[7,] 5801 17958 
[8,] 5801 16023 
[9,] 5801 7166 
[10,] 18451 17958 
[11,] 18451 16023 
[12,] 18451 7166 
    :  :  : 
    :  :  :

來源

2017-08-21 06:44:30 Onyambu

您已經有置換的方案。要打破數組併合並它，請逐行打開csv讀取並追加到列表中。

from itertools import chain 
import itertools 
#Create Empty Dictionary 
list= [] 
for i, eline in enumerate(CSVfile.readlines()): 
    list.append(eline.strip()) 
MergedArray= {i for j in (list) for i in j} 
#Use your permutations code below 
print list(itertools.permutations(MergedArray, 2))

來源

2017-08-21 07:01:30

每行每個項目的迭代置換

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