'How to get a value from a cell of a dataframe?
I have constructed a condition that extract 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?
Solution 1:[1]
These are fast access 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
Solution 2:[2]
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]
Solution 3:[3]
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
Solution 4:[4]
It doesn't need to be complicated:
val = df.loc[df.wd==1, 'col_name'].values[0]
Solution 5:[5]
I needed the value of one cell, selected by column and index names. This solution worked for me:
original_conversion_frequency.loc[1,:].values[0]
Solution 6:[6]
It looks like changes after pandas 10.1/13.1
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
Solution 7:[7]
The quickest/easiest options I have found are the following. 501 represents the row index.
df.at[501,'column_name']
df.get_value(501,'column_name')
Solution 8:[8]
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
Solution 9:[9]
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
Solution 10:[10]
For pandas 0.10, where iloc is unavalable, 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 then 1 row filtered, obtain the first row value. There will be an exception if the filter result in empty data frame.
Solution 11:[11]
Converting it to integer worked for me:
int(sub_df.iloc[0])
Solution 12:[12]
I've run across this when using DataFrames with MultiIndexes and found squeeze useful.
From the docs:
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 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 6-level index and 2-level columns, so only having to specify the outer level is quite helpful.
Solution 13:[13]
This is quite old by now but as of today you can fix it by simply doing
val = float(d2['col_name'].iloc[0])
Solution 14:[14]
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
Solution 15:[15]
To get the full row's value as JSON (instead of a Serie):
row = df.iloc[0]
Use the to_json method like bellow:
row.to_json()
Sources
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