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 dataframe:

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 dataframe 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?

Answer 1

You can get the values like this:

df[(df['column1']==any_value) & (df['column2']==any_value) & (df['column']==any_value)]['column_with_values_to_get']

And you can add (df['columnx']==any_value) as much as you want

Answer 2

If a single row was filtered from a dataframe, one way to get a scalar value from a single cell is squeeze() (or item()):

df = pd.DataFrame({'A':range(5), 'B': range(5)})
d2 = df[df['A'].le(5) & df['B'].eq(3)]
val = d2['A'].squeeze()                 # 3

val = d2['A'].item()                    # 3

In fact, item() may be called on the index, so item + at combo could work.

msk = df['A'].le(5) & df['B'].eq(3)
val = df.at[df.index[msk].item(), 'B']  # 3

In fact, the latter method is much faster than any other method listed here to get a single cell value.

df = pd.DataFrame({'A':range(10000), 'B': range(10000)})
msk = df['A'].le(5) & df['B'].eq(3)

%timeit df.at[df.index[msk].item(), 'A']
# 31.4 µs ± 5.83 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
%timeit df.loc[msk, 'A'].squeeze()
# 143 µs ± 8.99 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
%timeit df.loc[msk, 'A'].item()
# 125 µs ± 1.56 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
%timeit df.loc[msk, 'A'].iat[0]
# 125 µs ± 1.96 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
%timeit df[msk]['A'].values[0]
# 189 µs ± 8.67 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

Answer 3

Converting it to integer worked for me but if you need float it is also simple:

int(sub_df.iloc[0])

for float:

float(sub_df.iloc[0])

Answer 4

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]])

Answer 5

I needed the value of one cell, selected by column and index names. This solution worked for me:

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

Answer 6

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

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

Answer 7

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.

Answer 8

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

Answer 9

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()

Answer 10

You can turn your 1x1 dataframe into a NumPy array, then access the first and only value of that array:

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

Answer 11

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

Answer 12

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')

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

It doesn't need to be complicated:

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

Answer 19

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