CODEWITHSUNDEEP
pythonstringcsvsplit
Philipp
Philipp

Reputation: 333

Converting sublist of string (csv) to float (list comprehension)

I try to read a csv file which I need to convert into a list of floats for evaluation. The list looks like this:

[['Time [s];17063_X;17063_Y;17063_Z;17064_X;17064_Y;17064_Z;17065_X;17065_Y;17065_Z;17067_X;17067_Y;17067_Z;17068_X;17068_Y;17068_Z'], ['0;0.01952;0.04337;0.0242;0.01151;0.04152;0.03236;0.00015;-0.01679;0.05328;0.02872;0.01717;0.09341;0.01452;0.01489;0.07444'], ['0.00042;0.02188;0.04351;0.02803;0.0062;0.04108;0.03312;-0.00529;-0.01412;0.05167;0.02173;0.01377;0.04098;0.00807;0.00246;0.04354'],...]

but actually it has more than 17000 additional entries. The list I need should look like this:

[['Time [s]', '17063_X', '17063_Y', '17063_Z', '17064_X', '17064_Y', '17064_Z', '17065_X', '17065_Y', '17065_Z', '17067_X', '17067_Y', '17067_Z', '17068_X', '17068_Y', '17068_Z'], [0, 0.01952, 0.04337, 0.0242, 0.01151, 0.04152, 0.03236, 0.00015, -0.01679, 0.05328, 0.02872, 0.01717, 0.09341, 0.01452, 0.01489, 0.07444], ...]

So far I managed to get a single line (last element in list) into this format but not all the list. Here is what I've got so far:

import csv

with open(filepath, 'r') as f:
  reader = csv.reader(f)
  data = list(reader)

for j in range(1, len(data)):    # this loop does nothing?!
    for i in data[j]:
    dt = i.split(';')

da = [float(i) for i in dt]
print(da)

Out:

[0.005, 0.0207, 0.02925, 0.02095, 0.02332, 0.04211, 0.02223, 0.0075, -0.01961, 0.05093, 0.02604, 0.00711, 0.06644, 0.00689, -0.00092, 0.04737]

I would appreciate any help and also some tips when it comes to list comprehension. Thanks!

Upvotes: 2

Views: 292

Answers (5)

Aaditya Ura
Aaditya Ura

Reputation: 12689

You can simply try:

data=[['Time [s];17063_X;17063_Y;17063_Z;17064_X;17064_Y;17064_Z;17065_X;17065_Y;17065_Z;17067_X;17067_Y;17067_Z;17068_X;17068_Y;17068_Z'], ['0;0.01952;0.04337;0.0242;0.01151;0.04152;0.03236;0.00015;-0.01679;0.05328;0.02872;0.01717;0.09341;0.01452;0.01489;0.07444'], ['0.00042;0.02188;0.04351;0.02803;0.0062;0.04108;0.03312;-0.00529;-0.01412;0.05167;0.02173;0.01377;0.04098;0.00807;0.00246;0.04354']]


print(list(map(lambda x:list(map(lambda y:y.split(';'),x)),data)))

output:

[['Time [s]', '17063_X', '17063_Y', '17063_Z', '17064_X', '17064_Y', '17064_Z', '17065_X', '17065_Y', '17065_Z', '17067_X', '17067_Y', '17067_Z', '17068_X', '17068_Y', '17068_Z'], ['0', '0.01952', '0.04337', '0.0242', '0.01151', '0.04152', '0.03236', '0.00015', '-0.01679', '0.05328', '0.02872', '0.01717', '0.09341', '0.01452', '0.01489', '0.07444'], ['0.00042', '0.02188', '0.04351', '0.02803', '0.0062', '0.04108', '0.03312', '-0.00529', '-0.01412', '0.05167', '0.02173', '0.01377', '0.04098', '0.00807', '0.00246', '0.04354']]

Upvotes: 1

Oluwafemi Sule
Oluwafemi Sule

Reputation: 38992

Working with what you have. csv.reader can be passed a delimiter argument.

import csv

with open(filepath, 'r') as f:
  reader = csv.reader(f, delimiter=':')

reader supports one-time direct iteration of an open file handle within the context like so:

for data in reader:
    print(data)

If you need to use data in reader outside the context scope, bind a list containing its data to some other name inside the context scope.

import csv

entries = []
with open(filepath, 'r') as f:
  reader = csv.reader(f, delimiter=':')
  entries = list(reader)

Upvotes: 0

francisaugusto
francisaugusto

Reputation: 1197

Apparently, your csv file uses a ;, and not a ,.

I think you should try specifying the delimiter of your csv file. I haven't tested it, but you could do it like this:

import csv
with open(filepath, 'r') as f:
    reader = csv.reader(f, delimiter = ';')
    # omit the first line  
    for row in reader[1:]:
        da = [float(i) for i in dt]
        print (da)

Upvotes: 0

mij
mij

Reputation: 532

You need to put the last 2 lines inside the for loop, and also check your indentation:

for j in range(1, len(data)):
    for i in data[j]:
        dt = i.split(';')
        da = [float(v) for v in dt]
        print(da)

Upvotes: 1

jan-seins
jan-seins

Reputation: 1293

try the following 

s=[["a;b;c"],["d;e;f"]]
[x[0].split(';') for x in s]

and you will get:

[['a', 'b', 'c'], ['d', 'e', 'f']]

Upvotes: 0

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