python - Constructing pandas DataFrame from values in variables gives "ValueError: If using all scalar values, you must pass an index"

Question

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.

Answer 1

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

Answer 2

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')

Answer 3

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')

Answer 4

You may try wrapping your dictionary in to list

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

pd.DataFrame([my_dict])

   A  B
0  1  2

Answer 5

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 :

• Concatenate multiple Series into one data frame (as described here )

• Add a Series variable into existing data frame ( example here )

Answer 6

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)

Answer 7

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

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

Answer 8

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.

Answer 9

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’.

Answer 10

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

Answer 11

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)

Answer 12

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

Answer 13

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!

Answer 14

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]}

Answer 15

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']

Answer 16

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

Answer 17

simplest options ls :

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

Answer 18

I tried transpose() and it worked. Downside: You create a new object.

testdict1 = {'key1':'val1','key2':'val2','key3':'val3','key4':'val4'}

df = pd.DataFrame.from_dict(data=testdict1,orient='index')
print(df)
print(f'ID for DataFrame before Transpose: {id(df)}\n')

df = df.transpose()
print(df)
print(f'ID for DataFrame after Transpose: {id(df)}')

Output

         0
key1  val1
key2  val2
key3  val3
key4  val4
ID for DataFrame before Transpose: 1932797100424

   key1  key2  key3  key4
0  val1  val2  val3  val4
ID for DataFrame after Transpose: 1932797125448

​```

Answer 19

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

Answer 20

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()})

Answer 21

Just pass the dict on a list:

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