Write dictionary of lists to a CSV file

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

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)

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

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))