# How can I get a value from a cell of a dataframe?

## Question:

I have constructed a condition that extracts exactly one row from my data frame:

```
d2 = df[(df['l_ext']==l_ext) & (df['item']==item) & (df['wn']==wn) & (df['wd']==1)]
```

Now I would like to take a value from a particular column:

```
val = d2['col_name']
```

But as a result, I get a data frame that contains one row and one column (i.e., one cell). It is not what I need. I need one value (one float number). How can I do it in pandas?

## Answers:

If you have a DataFrame with only one row, then access the first (only) row as a Series using *iloc*, and then the value using the column name:

```
In [3]: sub_df
Out[3]:
A B
2 -0.133653 -0.030854
In [4]: sub_df.iloc[0]
Out[4]:
A -0.133653
B -0.030854
Name: 2, dtype: float64
In [5]: sub_df.iloc[0]['A']
Out[5]: -0.13365288513107493
```

These are fast access methods for scalars:

```
In [15]: df = pandas.DataFrame(numpy.random.randn(5, 3), columns=list('ABC'))
In [16]: df
Out[16]:
A B C
0 -0.074172 -0.090626 0.038272
1 -0.128545 0.762088 -0.714816
2 0.201498 -0.734963 0.558397
3 1.563307 -1.186415 0.848246
4 0.205171 0.962514 0.037709
In [17]: df.iat[0, 0]
Out[17]: -0.074171888537611502
In [18]: df.at[0, 'A']
Out[18]: -0.074171888537611502
```

It looks like changes after pandas 10.1 or 13.1.

I upgraded from 10.1 to 13.1. Before, *iloc* is not available.

Now with 13.1, `iloc[0]['label']`

gets a single value array rather than a scalar.

Like this:

```
lastprice = stock.iloc[-1]['Close']
```

Output:

```
date
2014-02-26 118.2
name:Close, dtype: float64
```

For pandas 0.10, where iloc is unavailable, filter a `DF`

and get the first row data for the column `VALUE`

:

```
df_filt = df[df['C1'] == C1val & df['C2'] == C2val]
result = df_filt.get_value(df_filt.index[0],'VALUE')
```

If there is more than one row filtered, obtain the first row value. There will be an exception if the filter results in an empty data frame.

I am not sure if this is a good practice, but I noticed I can also get just the value by casting the series as `float`

.

E.g.,

```
rate
```

3 0.042679

Name: Unemployment_rate, dtype: float64

```
float(rate)
```

0.0426789

The quickest and easiest options I have found are the following. *501* represents the row index.

```
df.at[501, 'column_name']
df.get_value(501, 'column_name')
```

Most answers are using *iloc* which is good for selection by position.

If you need selection-by-label, *loc* would be more convenient.

For getting a value explicitly (equiv to deprecated

df.get_value(‘a’,’A’))`# This is also equivalent to df1.at['a','A'] In [55]: df1.loc['a', 'A'] Out[55]: 0.13200317033032932`

You can turn your 1×1 dataframe into a NumPy array, then access the first and only value of that array:

```
val = d2['col_name'].values[0]
```

```
df_gdp.columns
```

Index([u’Country’, u’Country Code’, u’Indicator Name’, u’Indicator Code’,

u’1960′, u’1961′, u’1962′, u’1963′, u’1964′, u’1965′, u’1966′, u’1967′,

u’1968′, u’1969′, u’1970′, u’1971′, u’1972′, u’1973′, u’1974′, u’1975′,

u’1976′, u’1977′, u’1978′, u’1979′, u’1980′, u’1981′, u’1982′, u’1983′,

u’1984′, u’1985′, u’1986′, u’1987′, u’1988′, u’1989′, u’1990′, u’1991′,

u’1992′, u’1993′, u’1994′, u’1995′, u’1996′, u’1997′, u’1998′, u’1999′,

u’2000′, u’2001′, u’2002′, u’2003′, u’2004′, u’2005′, u’2006′, u’2007′,

u’2008′, u’2009′, u’2010′, u’2011′, u’2012′, u’2013′, u’2014′, u’2015′,

u’2016′],

dtype=’object’)

```
df_gdp[df_gdp["Country Code"] == "USA"]["1996"].values[0]
```

8100000000000.0

I needed the value of one cell, selected by column and index names.

This solution worked for me:

`df.loc[1,:].values[0]`

It doesn’t need to be complicated:

```
val = df.loc[df.wd==1, 'col_name'].values[0]
```

To get the full row’s value as JSON (instead of a Serie):

```
row = df.iloc[0]
```

Use the `to_json`

method like below:

```
row.to_json()
```

Converting it to integer worked for me:

```
int(sub_df.iloc[0])
```

I’ve run across this when using dataframes with MultiIndexes and found squeeze useful.

From the documentation:

Squeeze 1 dimensional axis objects into scalars.

Series or DataFrames with a single element are squeezed to a scalar.

DataFrames with a single column or a single row are squeezed to a

Series. Otherwise the object is unchanged.

```
# Example for a dataframe with MultiIndex
> import pandas as pd
> df = pd.DataFrame(
[
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
],
index=pd.MultiIndex.from_tuples( [('i', 1), ('ii', 2), ('iii', 3)] ),
columns=pd.MultiIndex.from_tuples( [('A', 'a'), ('B', 'b'), ('C', 'c')] )
)
> df
A B C
a b c
i 1 1 2 3
ii 2 4 5 6
iii 3 7 8 9
> df.loc['ii', 'B']
b
2 5
> df.loc['ii', 'B'].squeeze()
5
```

Note that while `df.at[]`

also works (if you aren’t needing to use conditionals) you then still AFAIK need to specify all levels of the MultiIndex.

Example:

```
> df.at[('ii', 2), ('B', 'b')]
5
```

I have a dataframe with a six-level index and two-level columns, so only having to specify the outer level is quite helpful.

In later versions, you can fix it by simply doing:

```
val = float(d2['col_name'].iloc[0])
```

Using `.item()`

returns a scalar (not a `Series`

), and it **only** works if there is a single element selected. It’s much safer than `.values[0]`

which will return the first element regardless of how many are selected.

```
>>> df = pd.DataFrame({'a': [1,2,2], 'b': [4,5,6]})
>>> df[df['a'] == 1]['a'] # Returns a Series
0 1
Name: a, dtype: int64
>>> df[df['a'] == 1]['a'].item()
1
>>> df2 = df[df['a'] == 2]
>>> df2['b']
1 5
2 6
Name: b, dtype: int64
>>> df2['b'].values[0]
5
>>> df2['b'].item()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3/dist-packages/pandas/core/base.py", line 331, in item
raise ValueError("can only convert an array of size 1 to a Python scalar")
ValueError: can only convert an array of size 1 to a Python scalar
```

**Display the data from a certain cell in pandas dataframe**

*Using dataframe.iloc*,

Dataframe.iloc should be used when given index is the actual index made when the pandas dataframe is created.

Avoid using dataframe.iloc on custom indices.

```
print(df['REVIEWLIST'].iloc[df.index[1]])
```

*Using dataframe.loc*,

Use dataframe.loc if you’re using a custom index it can also be used instead of iloc too even the dataframe contains default indices.

```
print(df['REVIEWLIST'].loc[df.index[1315]])
```