Please clear the Doubt

Question

#1
import pandas as pd
d={'a': 10, 'b': 56}
df=pd.DataFrame(d, index=[2])
print (df)

#2
import pandas as pd
d={'a': 10, 'b': 56}
df=pd.DataFrame(d)
print (df)

1st program gives a valid output while the 2nd program results in an error. Why is that?

Answer 1

In the first program, the DataFrame is created by passing a dictionary `d` to `pd.DataFrame()` along with specifying the `index=[2]`. This means the DataFrame will have only one row with the specified index '2', and the columns will be 'a' and 'b' with corresponding values 10 and 56, respectively. The output will look like this:

```
a b
2 10 56
```

In the second program, the DataFrame is created by passing the dictionary `d` to `pd.DataFrame()` without specifying an index. By default, pandas will use integer indices starting from 0. Since the dictionary `d` does not have any nested lists or arrays, pandas tries to interpret it as a Series, but the provided dictionary is not in the expected format for creating a Series. As a result, it raises an error.

If you want the second program to work correctly, you should either pass a list of dictionaries to create multiple rows or convert the dictionary to a list of tuples before creating the DataFrame:

Answer 2

I changed the 2nd code line to:
d={'a': [10], 'b': [56]}
And it works.
You can read more here:
https://www.geeksforgeeks.org/JUMP_LINK__&&__python__&&__JUMP_LINK-pandas-dataframe/

Answer 3

Ray ,

(ray - ? why the index in the first sample is set to ... index = [2] ?)

> the solution already suggested means that we have to modify possibly a big number of values by surrounding all of them with squared brackets.

this can be avoided by using the brackets around the dict like:

...
df=pd.DataFrame([d])
... ^ ^

so both versions should now give the same result in output.
https://code.sololearn.com/c3u5Tyf1rJLH/?ref=app

Answer 4

‎ ຸ ,

the code of *option 2* will not create the same output as *option 1*:

(option 1)
a b
0 10 56

(option 2)
col_name col_value
0 a 10
1 b 56

Answer 5

Option 1 (list of dictionaries):import pandas as pd
data = [{'a': 10, 'b': 56}]
df = pd.DataFrame(data)
print(df)

Option 2 (list of tuples):import pandas as pd
data = [('a', 10), ('b', 56)]
df = pd.DataFrame(data, columns=['col_name', 'col_value'])
print(df)

Both options will create the following DataFrame:a b
0 10 56

Answer 6

Lothar The difference between the two programs lies in the way the data is provided to create the DataFrame.

In the first program, you are providing a dictionary `d` with keys 'a' and 'b', and a single index value `[2]`. When you create the DataFrame using `pd.DataFrame(d, index=[2])`, it assigns the values 10 and 56 to the columns 'a' and 'b', respectively, and sets the index to 2. The output will look like this:

```
a b
2 10 56
```

In the second program, you are providing the same dictionary `d`, but without specifying any index. When you create the DataFrame using `pd.DataFrame(d)`, it treats the keys 'a' and 'b' as the column names and attempts to infer the index from the available data. Since no index is provided, it will use the default integer index range `[0, 1]`. The output will look like this:

```
a b
0 10 56
```

Both programs are valid, but the second one does not result in an error. It just generates a DataFrame with the default integer index. If you meant that you encountered

Answer 7

The error is raised because the dictionary d contains scalar values (10 and 56) instead of an iterable data structure like a list. When creating a DataFrame from a dictionary of scalar values, pandas expects you to specify an index. Since you didn't provide an index explicitly, it throws the ValueError because it cannot automatically infer the index from the scalar data.

To resolve this issue, you need to specify an index explicitly, even if you want to use the default integer index. You can do this by providing a list for the index, as shown in the first program:

python
Copy code
import pandas as pd
d = {'a': 10, 'b': 56}
df = pd.DataFrame(d, index=[0])
print(df)

Answer 8

Lothar ,

But in the 1st program, it worked without any square brackets around 'd'...
df=pd.DataFrame(d, index=[0])

What is the concept behind putting square brackets ?