Python：如何將列表與2D數組進行比較？

Question

我是數組主題的新手，並且在這種情況下使用for循環，因此我希望有人可以為我提供有關如何解決此問題的指導。

我有一個列表列表，看起來像這樣：

[[1, 0, 0], [0, 1, 0], [1, 1, 0], [0, 0, 1], [1, 0, 1], [0, 1, 1], [1, 1, 1]]

和一個二維數組，如下所示：

[[1 0 0]
 [1 0 1]
 [1 1 1]
 [1 0 0]
 [1 1 0]
 [1 0 1]
 [1 1 1]
 [0 0 1]
 [0 0 1]]

該數組最終將有近420k條記錄，我想統計一下我可以在列表列表中看到多少次組合。 我試過像這樣使用for循環：

from matplotlib import pyplot as plt
import numpy as np
from matplotlib_venn import venn3, venn3_circles
import os
import sys
from itertools import islice

input_file= "/home/ruchik/bodyMap_Data/bodyMap_Files/final.txt";
col0_idx = 6
col1_idx = 1
col2_idx = 2

print "number of sys arg", len(sys.argv)
print "sys arg list", sys.argv
input_file = sys.argv[1]
col1_idx = int(sys.argv[2])
col2_idx = int(sys.argv[3])

## keep it real ;)
col1_idx -= 1
col2_idx -= 1


print >> sys.stderr,'File is {file} and used columns are {col1} and {col2}.'.format(file=input_file, col1=col0_idx+col1_idx+1, col2=col2_idx+col0_idx+1)

## Openning and reading the file
#f = open(input_file, "r")
#g = open("final_fixed.txt", "w")
#print "Opened File Handle", f
#
#for line in f:
#    if line.strip():
#        g.write("\t".join(line.split()[7:]) + "\n")

#f.close()
#g.close()

#print "File created."

#f = open("final_fixed.txt", "r")
f = open(input_file, "r")

# header_all is a list of the content of the 1st line from position col0_idx-th to last-column-th
header_all_list = []
header_all_list = f.readline().rstrip("\n").split('\t')[col0_idx:]
header_reduced = [header_all_list[col1_idx], header_all_list[col2_idx], 'others']



# data_all is a (line-wise) list of (column-wise) list 
# with the content of each line but the 1st one from position col0_idx-th to last-column-th
data_all_lol = []
for line in f:
        data_all_lol.append(line.rstrip("\n").split('\t')[col0_idx:])

# just print the data_all list of list ... to make sure it is all fine up to there
#for i in range(len(data_all)):
#        for j in range(len(data_all[i])):
#                print >> sys.stderr, 'all data {col_i} , {col_j} : {val_ij}'.format(col_i=i+1, col_j=j+1+col0_idx, val_ij = data_all[i][j])

op_lol = [[1, 0, 0], [0, 1, 0], [1, 1, 0], [0, 0, 1], [1, 0, 1], [0, 1, 1], [1, 1, 1]]
count = [0, ] * len(op_lol)
for i in range(len(op_lol)):
    for j in range(len(data_reduced_transposed_npa)):
        if list(data_reduced_transposed_npa[j]) == op_lol[i]:
            count[i] += 1

op = [[1, 0, 0], [0, 1, 0], [1, 1, 0], [0, 0, 1], [1, 0, 1], [0, 1, 1], [1, 1, 1]]
count2 = [0, ] * len(op_lol)
for column in data_reduced_npa:
        for j in range(len(op_lol)):
                count2[j] += 1

#for k in range(len(op)):
#    print str(op[k]) + ': ' + str(count[k])
#header_venn3 = header_venn
#data_venn3 = data_venn
#print >> sys.stderr,"\nvenn3  order :"
#print >> sys.stderr,"(Abc, aBc, ABc, abC, AbC, aBC, ABC)"
#print >> sys.stderr,"venn3  header :"
#print >> sys.stderr, header_venn3
#print >> sys.stderr,"venn3 data :"
#for i in range(len(data_venn3)):
#    for j in range(len(data_venn3[i])):
#        print >> sys.stderr, 'venn3 data {col_i} , {col_j} : {val_ij}'.format(col_i=i, col_j=j, val_ij = data_venn3[i][j])




## Making the venn'
plt.figure(figsize=(4,4))
v = venn3(subsets=count, set_labels = ('Introns', 'Heart', 'Others'))
v.get_patch_by_id('100').set_alpha(1.0)
v.get_patch_by_id('100').set_color('white')
v.get_label_by_id('100').set_text('Unknown')
plt.show()

但這只是轉置2D陣列並為我打印，我在做什么錯？

Answer 1

您不能這樣做：

count = [0, ] * len(op_lol);

或這個：

count2 = [0, ] * len(op_lol);

這些正在內存中創建值0的淺表副本，因此當您索引這些count列表並為其分配新值時，您僅覆蓋了內存中的單個位置。 您實際上需要通過調用for循環，使用range ，使用map或使用copy.deepcopy()方法來實例化列表。

另外，您也不會說data_reduced_npa和data_reduced_transposed_npa來自何處或它們是什么，所以不能說是什么導致了您的輸出。 但是至少您應該研究一下copy.deepcopy() ：

https://docs.python.org/2/library/copy.html