Write dictionary of lists to a CSV file

Question

I'm struggling with writing a dictionary of lists to a .csv file.

This is how my dictionary looks like:

dict[key1]=[1,2,3]
dict[key2]=[4,5,6]
dict[key3]=[7,8,9]

I want the .csv file to look like:

key1  key2  key3
1     4     7  
2     5     8
3     6     9

At first I write the header:

outputfile = open (file.csv,'wb')
writefile = csv.writer (outputfile)
writefile.writerow(dict.keys())

So far so good... However, my problem is that I don't know how I could assign one list to the corresponding column. e.g.:

for i in range(0,len(dict[key1])):
    writefile.writerow([dict[key1][i],dict[key2][i],dict[key3][i])

will randomly fill the columns. Another problem is, that I have to manually fill in the keys and can't use it for another dictionary with 4 keys.

@RobWatts I think OP means that since dicts are unordered, the function keys() will print them out "randomly", eg. Key 3, Key 1, Key 2 — Cory Kramer
– Cory Kramer, Commented May 12, 2014 at 15:55

Tim Pietzcker · Accepted Answer · 2014-05-12 16:02:03Z

50

If you don't care about the order of your columns (since dictionaries are unordered), you can simply use zip():

d = {"key1": [1,2,3], "key2": [4,5,6], "key3": [7,8,9]}
with open("test.csv", "wb") as outfile:
   writer = csv.writer(outfile)
   writer.writerow(d.keys())
   writer.writerows(zip(*d.values()))

Result:

key3    key2    key1
7       4       1
8       5       2
9       6       3

If you do care about order, you need to sort the keys:

keys = sorted(d.keys())
with open("test.csv", "wb") as outfile:
   writer = csv.writer(outfile, delimiter = "\t")
   writer.writerow(keys)
   writer.writerows(zip(*[d[key] for key in keys]))

Result:

key1    key2    key3
1       4       7
2       5       8
3       6       9

edited May 12, 2014 at 16:02

answered May 12, 2014 at 15:56

Tim Pietzcker

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9 Comments

Chit Khine Over a year ago

how can we change rows to columns in here key1 \tab 7 \tab 4 \tab 1

Tim Pietzcker Over a year ago

If you look at the question, I used zip() to transpose the original data. If you do the operation again, you'll change rows to columns again. Note that key1 - 7 - 4 - 1 is probably not what you want...

jwoe Over a year ago

This runs for me and I get the output I want, but it still throws a typeError "zip argument #1 must support iteration". Any reason for this?

Tim Pietzcker Over a year ago

Right, on Python 2, you need to use wb, whereas on Python 3, you need to use the newline='' option when opening a CSV file.

Yaboku Over a year ago

@JasonGoal it maybe very late but if you are still wondering about the answer here is a gist where the person explained it perfectly gist.github.com/captmomo/7836eb53b7cee9fb7d38460f70942902

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Kumar Shubham · Accepted Answer · 2019-08-01 12:33:46Z

4

This will work even when the list in key are of different length.

    with myFile:  
        writer = csv.DictWriter(myFile, fieldnames=list(clusterWordMap.keys()))   
        writer.writeheader()
        while True:
            data={}
            for key in clusterWordMap:
                try:
                    data[key] = clusterWordMap[key][ind]
                except:
                    pass
            if not data:
                break
            writer.writerow(data)

You can use pandas for saving it into csv:

df = pd.DataFrame({key: pd.Series(value) for key, value in dictmap.items()})
df.to_csv(filename, encoding='utf-8', index=False)

edited Aug 1, 2019 at 12:33

answered May 17, 2019 at 10:51

Kumar Shubham

393 bronze badges

3 Comments

Austin Over a year ago

df = pd.DataFrame(dict) is all you need to create the dataframe.

Kumar Shubham Over a year ago

it will fail if value stored in dictionary is of varying length

uoay Over a year ago

Could you define myFile, clusterWordMap, and ind? Thanks.

Alireza · Accepted Answer · 2018-04-23 20:19:59Z

3

save:

with open(path, 'a') as csv_file:
    writer = csv.writer(csv_file)
    for key, value in dict_.items():
        writer.writerow([key, ','.join(value)])
csv_file.close()        
print ('saving is complete')

read back:

with open(csv_path, 'rb') as csv_file:
    reader = csv.reader(csv_file);
    temp_dict = dict(reader);
mydict={k:v.split(',') for k,v in temp_dict.items()}    
csv_file.close()
return mydict

answered Apr 23, 2018 at 20:19

Alireza

311 bronze badge

Comments

jedwards · Accepted Answer · 2014-05-12 15:58:48Z

2

Given

dict = {}
dict['key1']=[1,2,3]
dict['key2']=[4,5,6]
dict['key3']=[7,8,9]

The following code:

COL_WIDTH = 6
FMT = "%%-%ds" % COL_WIDTH

keys = sorted(dict.keys())

with open('out.csv', 'w') as csv:
    # Write keys    
    csv.write(''.join([FMT % k for k in keys]) + '\n')

    # Assume all values of dict are equal
    for i in range(len(dict[keys[0]])):
        csv.write(''.join([FMT % dict[k][i] for k in keys]) + '\n')

produces a csv that looks like:

key1  key2  key3
1     4     7
2     5     8
3     6     9

answered May 12, 2014 at 15:58

jedwards

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Comments

wwii · Accepted Answer · 2014-05-12 17:43:40Z

2

Roll your own without the csv module:

d = {'key1' : [1,2,3],
     'key2' : [4,5,6],
     'key3' : [7,8,9]}

column_sequence = sorted(d.keys())
width = 6
fmt = '{{:<{}}}'.format(width)
fmt = fmt*len(column_sequence) + '\n'

output_rows = zip(*[d[key] for key in column_sequence])

with open('out.txt', 'wb') as f:
    f.write(fmt.format(*column_sequence))
    for row in output_rows:
        f.write(fmt.format(*row))

answered May 12, 2014 at 17:43

wwii

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Comments

jcfollower · Accepted Answer · 2014-05-12 15:58:29Z

1

key_list = my_dict.keys()    
limit = len(my_dict[key_list[0]])    

for index in range(limit):    
  writefile.writerow([my_dict[x][index] for x in key_list])

answered May 12, 2014 at 15:58

jcfollower

3,15822 silver badges26 bronze badges

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