Converting a dictionary with lists for values into a dataframe

Question

I spent a while looking through SO and seems I have a unique problem.

I have a dictionary that looks like the following:

dict={
    123: [2,4],
    234: [6,8],
    ...
}

I want to convert this dictionary that has lists for values into a 3 column data frame like the following:

time, value1, value2
123, 2, 4
234, 6, 8
...

I can run:

pandas.DataFrame(dict)

but this generates the following:

123, 234, ...
2, 6, ...
4, 8, ...

Probably a simple fix but I'm still picking up pandas

Answer 1

You can either preprocess the data as levi suggests, or you can transpose the data frame after creating it.

testdict={
    123: [2,4],
    234: [6,8],
    456: [10, 12]
}
df = pd.DataFrame(testdict)
df = df.transpose()

print(df)
#      0  1
# 123  2  4
# 234  6  8

Answer 2

It may be of interest to some that Roger Fan's pandas.DataFrame(dict) method is actually pretty slow if you have a ton of indices. The faster way is to just preprocess the data into separate lists and then create a DataFrame out of these lists. (Perhaps this was explained in levi's answer, but it is gone now.)

For example, consider this dictionary, dict1, where each value is a list. Specifically, dict1[i] = [ i*10, i*100] (for ease of checking the final dataframe).

keys = range(1000)
values = zip(np.arange(1000)*10, np.arange(1000)*100)
dict1 = dict(zip(keys, values))

It takes roughly 30 times as long with the pandas method. E.g.

t = time.time()
test1 = pd.DataFrame(dict1).transpose()
print time.time() - t

0.118762016296

versus:

t = time.time()
keys = []
list1 = []
list2 = []
for k in dict1:
    keys.append(k)
    list1.append(dict1[k][0])
    list2.append(dict1[k][1])
test2 = pd.DataFrame({'element1': list1, 'element2': list2}, index=keys)
print time.time() - t

0.00310587882996