## The Question :

*443 people think this question is useful*

This may be a simple question, but I can not figure out how to do this. Lets say that I have two variables as follows.

a = 2
b = 3

I want to construct a DataFrame from this:

df2 = pd.DataFrame({'A':a,'B':b})

This generates an error:

ValueError: If using all scalar values, you must pass an index

I tried this also:

df2 = (pd.DataFrame({'a':a,'b':b})).reset_index()

This gives the same error message.

*The Question Comments :*

## The Answer 1

*676 people think this answer is useful*

The error message says that if you’re passing scalar values, you have to pass an index. So you can either not use scalar values for the columns — e.g. use a list:

>>> df = pd.DataFrame({'A': [a], 'B': [b]})
>>> df
A B
0 2 3

or use scalar values and pass an index:

>>> df = pd.DataFrame({'A': a, 'B': b}, index=[0])
>>> df
A B
0 2 3

## The Answer 2

*79 people think this answer is useful*

You can also use `pd.DataFrame.from_records`

which is more convenient when you already have the dictionary in hand:

df = pd.DataFrame.from_records([{ 'A':a,'B':b }])

You can also set index, if you want, by:

df = pd.DataFrame.from_records([{ 'A':a,'B':b }], index='A')

## The Answer 3

*61 people think this answer is useful*

You need to create a pandas series first. The second step is to convert the pandas series to pandas dataframe.

import pandas as pd
data = {'a': 1, 'b': 2}
pd.Series(data).to_frame()

You can even provide a column name.

pd.Series(data).to_frame('ColumnName')

## The Answer 4

*45 people think this answer is useful*

You may try wrapping your dictionary in to list

`my_dict = {'A':1,'B':2}`

`pd.DataFrame([my_dict])`

A B
0 1 2

## The Answer 5

*13 people think this answer is useful*

Maybe Series would provide all the functions you need:

pd.Series({'A':a,'B':b})

DataFrame can be thought of as a collection of Series hence you can :

## The Answer 6

*8 people think this answer is useful*

You need to provide iterables as the values for the Pandas DataFrame columns:

df2 = pd.DataFrame({'A':[a],'B':[b]})

## The Answer 7

*8 people think this answer is useful*

I had the same problem with numpy arrays and the solution is to flatten them:

data = {
'b': array1.flatten(),
'a': array2.flatten(),
}
df = pd.DataFrame(data)

## The Answer 8

*8 people think this answer is useful*

Pandas magic at work. All logic is out.

The error message `"ValueError: If using all scalar values, you must pass an index"`

Says you must pass an index.

This does not necessarily mean passing an index makes pandas do **what you want it to do**

When you pass an index, pandas will treat your dictionary keys as column names and the values as what the column should contain for each of the values in the index.

a = 2
b = 3
df2 = pd.DataFrame({'A':a,'B':b}, index=[1])
A B
1 2 3

Passing a larger index:

df2 = pd.DataFrame({'A':a,'B':b}, index=[1, 2, 3, 4])
A B
1 2 3
2 2 3
3 2 3
4 2 3

An index is usually automatically generated by a dataframe when none is given. However, pandas does not know how many rows of `2`

and `3`

you want. You can however be more explicit about it

df2 = pd.DataFrame({'A':[a]*4,'B':[b]*4})
df2
A B
0 2 3
1 2 3
2 2 3
3 2 3

The default index is 0 based though.

I would recommend always passing a dictionary of lists to the dataframe constructor when creating dataframes. It’s easier to read for other developers. Pandas has a lot of caveats, don’t make other developers have to experts in all of them in order to read your code.

## The Answer 9

*6 people think this answer is useful*

You could try:

df2 = pd.DataFrame.from_dict({'a':a,'b':b}, orient = 'index')

From the documentation on the ‘orient’ argument: If the keys of the passed dict should be the columns of the resulting DataFrame, pass ‘columns’ (default). Otherwise if the keys should be rows, pass ‘index’.

## The Answer 10

*3 people think this answer is useful*

If you intend to convert a dictionary of scalars, you have to include an index:

import pandas as pd
alphabets = {'A': 'a', 'B': 'b'}
index = [0]
alphabets_df = pd.DataFrame(alphabets, index=index)
print(alphabets_df)

Although index is not required for a dictionary of lists, the same idea can be expanded to a dictionary of lists:

planets = {'planet': ['earth', 'mars', 'jupiter'], 'length_of_day': ['1', '1.03', '0.414']}
index = [0, 1, 2]
planets_df = pd.DataFrame(planets, index=index)
print(planets_df)

Of course, for the dictionary of lists, you can build the dataframe without an index:

planets_df = pd.DataFrame(planets)
print(planets_df)

## The Answer 11

*3 people think this answer is useful*

the input does not have to be a list of records – it can be a single dictionary as well:

pd.DataFrame.from_records({'a':1,'b':2}, index=[0])
a b
0 1 2

Which seems to be equivalent to:

pd.DataFrame({'a':1,'b':2}, index=[0])
a b
0 1 2

## The Answer 12

*2 people think this answer is useful*

This is because a DataFrame has two intuitive dimensions – the columns *and* the rows.

You are only specifying the columns using the dictionary keys.

If you only want to specify one dimensional data, use a Series!

## The Answer 13

*1 people think this answer is useful*

Convert Dictionary to Data Frame

col_dict_df = pd.Series(col_dict).to_frame('new_col').reset_index()

Give new name to Column

col_dict_df.columns = ['col1', 'col2']

## The Answer 14

*1 people think this answer is useful*

Change your ‘a’ and ‘b’ values to a list, as follows:

a = [2]
b = [3]

then execute the same code as follows:

df2 = pd.DataFrame({'A':a,'B':b})
df2

and you’ll get:

A B
0 2 3

## The Answer 15

*1 people think this answer is useful*

I usually use the following to to quickly create a small table from dicts.

Let’s say you have a dict where the keys are filenames and the values their corresponding filesizes, you could use the following code to put it into a DataFrame (notice the .items() call on the dict):

files = {'A.txt':12, 'B.txt':34, 'C.txt':56, 'D.txt':78}
filesFrame = pd.DataFrame(files.items(), columns=['filename','size'])
print(filesFrame)
filename size
0 A.txt 12
1 B.txt 34
2 C.txt 56
3 D.txt 78

## The Answer 16

*0 people think this answer is useful*

You could try this:
df2 = pd.DataFrame.from_dict({‘a’:a,’b’:b}, orient = ‘index’)

## The Answer 17

*0 people think this answer is useful*

simplest options ls :

dict = {'A':a,'B':b}
df = pd.DataFrame(dict, index = np.arange(1) )

## The Answer 18

*0 people think this answer is useful*

Another option is to convert the scalars into list on the fly using Dictionary Comprehension:

df = pd.DataFrame(data={k: [v] for k, v in mydict.items()})

The expression {…} creates a new dict whose values is a list of 1 element. such as :

In [20]: mydict
Out[20]: {'a': 1, 'b': 2}
In [21]: mydict2 = { k: [v] for k, v in mydict.items()}
In [22]: mydict2
Out[22]: {'a': [1], 'b': [2]}

## The Answer 19

*-2 people think this answer is useful*

If you have a dictionary you can turn it into a pandas data frame with the following line of code:

pd.DataFrame({"key": d.keys(), "value": d.values()})

## The Answer 20

*-2 people think this answer is useful*

Just pass the dict on a list:

a = 2
b = 3
df2 = pd.DataFrame([{'A':a,'B':b}])