Sortie 2 dim array 'liste de listes' en fichier texte en python

Question

Question simple - Je crée un tableau de deux dim (ddist = [[0]*d for _ in [0]*d]) en utilisant les listes dans le code ci-dessous, il affiche la distance en utilisant les données gis. pour prendre le résultat de mon tableau/liste et sortie dans un fichier texte en gardant la même structure N * N. J'ai utilisé la sortie des instructions d'impression dans le passé, mais pas une bonne solution dans ce cas

Je suis nouveau à python par la SAS

def match_bg(): 
    #as the name suggests this function will match the variations of blockgroups with grid travel time. Then output into two arras time and distance. 
    count = -1 
    countwo = -1 
    ctime = -1 
    ddist = [[0]*d for _ in [0]*d] #cratesan N*N array list 
    dtime = -1 


    while count < 10: 
     count = count +1 
     #j[count][7] = float(j[count][7]) 
     #j[count][6] = float(j[count][6]) 
     while countwo < d: 
      countwo = countwo+1 
      if count < 1: 
       #change values in bg file 
       j[countwo][7] = float(j[countwo][7]) 
       j[countwo][6] = float(j[countwo][6]) 




      #print j[count], j[countwo] 
      while ctime < RowsT: 

       #print ctime, lenth, t[ctime][0], count, countwo 
       ctime = ctime + 1 



       #takes both verations of big zone which should be end of the file and matches to travetime file - note 0 and 1 for t[] should be same for different files 
       if ((j[count][lenth-1] == t[ctime][0]) and (j[countwo][lenth-1] == t[ctime][1])) or ((j[countwo][lenth-1] == t[ctime][0]) and (j[count][lenth-1] == t[ctime][1])): 
        if t[ctime][0] != t[ctime][1]: 
         #jkdljf 
         x1=3963*j[count][7]*(math.pi/180) 
         x2=3963*j[countwo][7]*(math.pi/180) 

         y1=math.cos(j[count][6]*math.pi/180)*3963*j[count][7]*(math.pi/180) 
         y2=math.cos(j[countwo][6]*math.pi/180)*3963*j[countwo][7]*(math.pi/180) 

         dist=math.sqrt(pow((x1-x2), 2) + pow((y1-y2), 2)) 

         dtime = dist/t[ctime][11] 
         print countwo, count 
         ddist[count-1][countwo-1] = dist/t[ctime][lenth] 
         print dtime, "ajusted time", "not same grid" 

         print 
        elif j[count][5] != j[countwo][5]: 
         #ljdkjfs 
         x1=3963*j[count][7]*(math.pi/180) 
         x2=3963*j[countwo][7]*(math.pi/180) 

         y1=math.cos(j[count][6]*math.pi/180)*3963*j[count][7]*(math.pi/180) 
         y2=math.cos(j[countwo][6]*math.pi/180)*3963*j[countwo][7]*(math.pi/180) 

         dist=math.sqrt(pow((x1-x2), 2) + pow((y1-y2), 2)) # could change to calculation 

         dtime = (dist/.65)/(t[ctime][10]/60.0) 


         print dtime, dist, "not in the same bg", j[count], j[countwo], t[ctime] 

        elif j[count][5] == j[countwo][5]: 
         if t[count][7] < 3000000: 
          dtime = 3 
         elif t[count][7] < 20000000: 
          dtime = 8 
         else: 
          dtime = 12 
         print dtime, "same bg" 
         print t[ctime][0], t[ctime], 1, j[count], j[countwo] 
        else: 
         print "error is skip logic", j[count], j[countwo], t[ctime] 
        break 
       #elif (j[countwo][lenth-1] == t[ctime][0]) and (j[count][lenth-1] == t[ctime][1]): 
        #print t[ctime][0], t[ctime], 2, j[count], j[countwo] 
        #break 

      ctime = -1 

     countwo = -1 

Answer 1

que ce que vous pouvez sortir votre liste 2d (ou une liste 2d pour cette question).

with open(outfile, 'w') as file: 
    file.writelines('\t'.join(str(j) for j in i) + '\n' for i in top_list)