printing sorted columns based on values in python - python-2.7

Question

I have a CSV file with 11 columns. I am new to Python and trying to make a code that sorted column 2 alphabetically (string) and then prints the rows of the sorted list.
I only want the rows printed where column 1 == 1.
If column 1 == 0 then i want the script to ignore this row and continue on to the next line without printing it.
in_file = open('filename.csv', 'r')
dict = {}
#should print all rows where field [0] == 1
print('List of rows where column one is equal to zero in alphabetical order of column 2')
for line in in_file:
line = line.strip('\n')
fields = line.split(',')
c1 = fields[0]
c2 = fields[1]
c3 = fields[2]
c4 = fields[3]
c5 = fields[4]
c6 = fields[5]
c7 = fields[6]
c8 = fields[7]
c9 = fields[8]
c10 = fields[9]
c11 = fields[10]
if c2 not in dict:
dict[c2] = c1
if c3 not in dict:
dict[c3] = c1
if c4 not in dict:
dict[c4] = c1
if c5 not in dict:
dict[c5] = c1
if c6 not in dict:
dict[c6] = c1
if c7 not in dict:
dict[c7] = c1
if c8 not in dict:
dict[c8] = c1
if c9 not in dict:
dict[c9] = c1
if c10 not in dict:
dict[c10] = c1
if c11 not in dict:
dict[c11] = c1
name = dict.keys()
sorted_names = sorted(name)
for name in sorted_names:
c1 = dict[name]
rows = [row for row in in_file if row ['c1']!= 1]
for row in rows:
print(c3, c2, c4, c5)
in_file.close()

Answer 1

As you commented, the error pops up in the 'rows = [row for row in in_file if row ['c1']!= 1]' line. As far as i know, the error is really simple:
you input indices as strings, but they Always have to be integers, as the error says. for example: your index is ['c1'], but c1 is a variable. using [c1] would probably solve the problem.