pandas DataFrame assign with format - python

I'm trying to use assign to create a new column in a pandas DataFrame. I need to use something like str.format to have the new column be pieces of existing columns. For instance...
import pandas as pd
df = pd.DataFrame(np.random.randn(3, 3))
gives me...
0 1 2
0 -0.738703 -1.027115 1.129253
1 0.674314 0.525223 -0.371896
2 1.021304 0.169181 -0.884293
an assign for a totally new column works
# works
print(df.assign(c = "a"))
0 1 2 c
0 -0.738703 -1.027115 1.129253 a
1 0.674314 0.525223 -0.371896 a
2 1.021304 0.169181 -0.884293 a
But, if I want to use an existing column into a new column it seems like pandas is adding the whole existing frame into the new column.
# doesn't work
print(df.assign(c = "a{}b".format(df[0])))
0 1 2 \
0 -0.738703 -1.027115 1.129253
1 0.674314 0.525223 -0.371896
2 1.021304 0.169181 -0.884293
c
0 a0 -0.738703\n1 0.674314\n2 1.021304\n...
1 a0 -0.738703\n1 0.674314\n2 1.021304\n...
2 a0 -0.738703\n1 0.674314\n2 1.021304\n...
Thanks for the help.

In [131]: df.assign(c="a"+df[0].astype(str)+"b")
Out[131]:
0 1 2 c
0 0.833556 -0.106183 -0.910005 a0.833556419295b
1 -1.487825 1.173338 1.650466 a-1.48782514804b
2 -0.836795 -1.192674 -0.212900 a-0.836795026809b
'a{}b'.format(df[0]) is a str. "a"+df[0].astype(str)+"b" is a Series.
In [142]: type(df[0].astype(str))
Out[142]: pandas.core.series.Series
In [143]: type('{}'.format(df[0]))
Out[143]: str
When you assign a single string to the column c, that string is repeated for every row in df.
Thus, df.assign(c = "a{}b".format(df[0])) assigns the string 'a{}b'.format(df[0])
to each row of df:
In [138]: 'a{}b'.format(df[0])
Out[138]: 'a0 0.833556\n1 -1.487825\n2 -0.836795\nName: 0, dtype: float64b'
It is really no different than what happened with df.assign(c = "a").
In contrast, when you assign a Series to the column c, then the index of the Series is aligned with the index of df and the corresponding values are assigned to df['c'].
Under the hood, the Series.__add__ method is defined in such a way so that addition of the Series containing strings with a string results in a new Series with the string concatenated with the values in the Series:
In [149]: "a"+df[0].astype(str)
Out[149]:
0 a0.833556419295
1 a-1.48782514804
2 a-0.836795026809
Name: 0, dtype: object
(The astype method was called to convert the floats in df[0] into strings.)

df['c'] = "a" + df[0].astype(str) + 'b'
df
0 1 2 c
0 -1.134154 -0.367397 0.906239 a-1.13415403091b
1 0.551997 -0.160217 -0.869291 a0.551996920472b
2 0.490102 -1.151301 0.541888 a0.490101854737b

Related

How can I assign a new column to a slice of a pandas DataFrame with a multiindex?

I have a pandas DataFrame with a multi-index like this:
import pandas as pd
import numpy as np
arr = [1]*3 + [2]*3
arr2 = list(range(3)) + list(range(3))
mux = pd.MultiIndex.from_arrays([
arr,
arr2
], names=['one', 'two'])
df = pd.DataFrame({'a': np.arange(len(mux))}, mux)
df
a
one two
1 0 0
1 1 1
1 2 2
2 0 3
2 1 4
2 2 5
I have a function that takes a slice of a DataFrame and needs to assign a new column to the rows that have been sliced:
def work(df):
b = df.copy()
#do some work on the slice and create values for a new column of the slice
b['b'] = b['a']*2
#assign the new values back to the slice in a new column
df['b'] = b['b']
#pass in a slice of the df with only records that have the last value for 'two'
work(df.loc[df.index.isin(df.index.get_level_values('two')[-1:], level=1)])
However calling the function results in the error:
SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
# This is added back by InteractiveShellApp.init_path()
How can I create a new column 'b' in the original DataFrame and assign its values for only the rows that were passed to the function, leaving the rest of the rows nan?
The desired output is:
a b
one two
1 0 0 nan
1 1 1 nan
1 2 2 4
2 0 3 nan
2 1 4 nan
2 2 5 10
NOTE: In the work function I'm actually doing a bunch of complex operations involving calling other functions to generate the values for the new column so I don't think this will work. Multiplying by 2 in my example is just for illustrative purposes.
You actually don't have an error, but just a warning. Try this:
def work(df):
b = df.copy()
#do some work on the slice and create values for a new column of the slice
b['b'] = b['a']*2
#assign the new values back to the slice in a new column
df['b'] = b['b']
return df
#pass in a slice of the df with only records that have the last value for 'two'
new_df = work(df.loc[df.index.isin(df.index.get_level_values('two')[-1:], level=1)])
Then:
df.reset_index().merge(new_df, how="left").set_index(["one","two"])
Output:
a b
one two
1 0 0 NaN
1 1 NaN
2 2 4.0
2 0 3 NaN
1 4 NaN
2 5 10.0
I don't think you need a separate function at all. Try this...
df['b'] = df['a'].where(df.index.isin(df.index.get_level_values('two')[-1:], level=1))*2
The Series.where() function being called on df['a'] here should return a series where values are NaN for rows that do not result from your query.

Added column to existing dataframe but entered all numbers as NaN

So I created two dataframes from existing CSV files, both consisting of entirely numbers. The second dataframe consists of an index from 0 to 8783 and one column of numbers and I want to add it on as a new column to the first dataframe which has an index consisting of a month, day and hour. I tried using append, merge and concat and none worked and then tried simply using:
x1GBaverage['Power'] = x2_cut
where x1GBaverage is the first dataframe and x2_cut is the second. When I did this it added x2_cut on properly but all the values were entered as NaN instead of the numerical values that they should be. How should I be approaching this?
x1GBaverage['Power'] = x2_cut.values
problem solved :)
The thing about pandas is that values are implicitly linked to their indices unless you deliberately specify that you only need the values to be transferred over.
If they're the same row counts and you just want to tack it on the end, the indexes either need to match, or you need to just pass the underlying values. In the example below, columns 3 and 5 are the index matching & value versions, and 4 is what you're running into now:
In [58]: df = pd.DataFrame(np.random.random((3,3)))
In [59]: df
Out[59]:
0 1 2
0 0.670812 0.500688 0.136661
1 0.185841 0.239175 0.542369
2 0.351280 0.451193 0.436108
In [61]: df2 = pd.DataFrame(np.random.random((3,1)))
In [62]: df2
Out[62]:
0
0 0.638216
1 0.477159
2 0.205981
In [64]: df[3] = df2
In [66]: df.index = ['a', 'b', 'c']
In [68]: df[4] = df2
In [70]: df[5] = df2.values
In [71]: df
Out[71]:
0 1 2 3 4 5
a 0.670812 0.500688 0.136661 0.638216 NaN 0.638216
b 0.185841 0.239175 0.542369 0.477159 NaN 0.477159
c 0.351280 0.451193 0.436108 0.205981 NaN 0.205981
If the row counts differ, you'll need to use df.merge and let it know which columns it should be using to join the two frames.

Pandas, selecting by column and row

I want to sum up all values that I select based on some function of column and row.
Another way of putting it is that I want to use a function of the row index and column index to determine if a value should be included in a sum along an axis.
Is there an easy way of doing this?
Columns can be selected using the syntax dataframe[<list of columns>]. The index (row) can be used for filtering using the dataframe.index method.
import pandas as pd
df = pd.DataFrame({'a': [0.1, 0.2], 'b': [0.2, 0.1]})
odd_a = df['a'][df.index % 2 == 1]
even_b = df['b'][df.index % 2 == 0]
# odd_a:
# 1 0.2
# Name: a, dtype: float64
# even_b:
# 0 0.2
# Name: b, dtype: float64
If df is your dataframe :
In [477]: df
Out[477]:
A s2 B
0 1 5 5
1 2 3 5
2 4 5 5
You can access the odd rows like this :
In [478]: df.loc[1::2]
Out[478]:
A s2 B
1 2 3 5
and the even ones like this:
In [479]: df.loc[::2]
Out[479]:
A s2 B
0 1 5 5
2 4 5 5
To answer your question, getting even rows and column B would be :
In [480]: df.loc[::2,'B']
Out[480]:
0 5
2 5
Name: B, dtype: int64
and odd rows and column A can be done as:
In [481]: df.loc[1::2,'A']
Out[481]:
1 2
Name: A, dtype: int64
I think this should be fairly general if not the cleanest implementation. This should allow applying separate functions for rows and columns depending on conditions (that I defined here in dictionaries).
import numpy as np
import pandas as pd
ran = np.random.randint(0,10,size=(5,5))
df = pd.DataFrame(ran,columns = ["a","b","c","d","e"])
# A dictionary to define what function is passed
d_col = {"high":["a","c","e"], "low":["b","d"]}
d_row = {"high":[1,2,3], "low":[0,4]}
# Generate list of Pandas boolean Series
i_col = [df[i].apply(lambda x: x>5) if i in d_col["high"] else df[i].apply(lambda x: x<5) for i in df.columns]
# Pass the series as a matrix
df = df[pd.concat(i_col,axis=1)]
# Now do this again for rows
i_row = [df.T[i].apply(lambda x: x>5) if i in d_row["high"] else df.T[i].apply(lambda x: x<5) for i in df.T.columns]
# Return back the DataFrame in original shape
df = df.T[pd.concat(i_row,axis=1)].T
# Perform the final operation such as sum on the returned DataFrame
print(df.sum().sum())

Grouping data in Python with pandas yields a blank first row

I have this nice pandas dataframe:
And I want to group it by the column "0" (which represents the year) and calculate the mean of the other columns for each year. I do such thing with this code:
df.groupby(0)[2,3,4].mean()
And that successfully calculates the mean of every column. The problem here being the empty row that appears on top:
That's just a display thing, the grouped column now becomes the index and this is just the way that it is displayed, you will notice here that even when you set pd.set_option('display.notebook_repr_html', False) you still get this line, it has no effect on operations on the goruped df:
In [30]:
df = pd.DataFrame({'a':np.random.randn(5), 'b':np.random.randn(5), 'c':np.arange(5)})
df
Out[30]:
a b c
0 0.766706 -0.575700 0
1 0.594797 -0.966856 1
2 1.852405 1.003855 2
3 -0.919870 -1.089215 3
4 -0.647769 -0.541440 4
In [31]:
df.groupby('c')['a','b'].mean()
Out[31]:
a b
c
0 0.766706 -0.575700
1 0.594797 -0.966856
2 1.852405 1.003855
3 -0.919870 -1.089215
4 -0.647769 -0.541440
Technically speaking it has assigneed the name attribute:
In [32]:
df.groupby('c')['a','b'].mean().index.name
Out[32]:
'c'
by default there will be no name if it has not been assigned:
In [34]:
print(df.index.name)
None

Pandas, concat Series to DF as rows

I attempting to add a Series to an empty DataFrame and can not find an answer
either in the Doc's or other questions. Since you can append two DataFrames by row
or by column it would seem there must be an "axis marker" missing from a Series. Can
anyone explain why this does not work?.
import Pandas as pd
df1 = pd.DataFrame()
s1 = pd.Series(['a',5,6])
df1 = pd.concat([df1,s1],axis = 1)
#go run some process return s2, s3, sn ...
s2 = pd.Series(['b',8,9])
df1 = pd.concat([df1,s2],axis = 1)
s3 = pd.Series(['c',10,11])
df1 = pd.concat([df1,s3],axis = 1)
If my example above is some how misleading perhaps using the example from the docs will help.
Quoting: Appending rows to a DataFrame.
While not especially efficient (since a new object must be created), you can append a
single row to a DataFrame by passing a Series or dict to append, which returns a new DataFrame as above. End Quote.
The example from the docs appends "S", which is a row from a DataFrame, "S1" is a Series
and attempting to append "S1" produces an error. My question is WHY will appending "S1 not work? The assumption behind the question is that a DataFrame must code or contain axes information for two axes, where a Series must contain only information for one axes.
df = pd.DataFrame(np.random.randn(8, 4), columns=['A','B','C','D'])
s = df.xs(3); #third row of DataFrame
s1 = pd.Series([np.random.randn(4)]); #new Series of equal len
df= df.append(s, ignore_index=True)
Result
0 1
0 a b
1 5 8
2 6 9
Desired
0 1 2
0 a 5 6
1 b 8 9
You were close, just transposed the result from concat
In [14]: s1
Out[14]:
0 a
1 5
2 6
dtype: object
In [15]: s2
Out[15]:
0 b
1 8
2 9
dtype: object
In [16]: pd.concat([s1, s2], axis=1).T
Out[16]:
0 1 2
0 a 5 6
1 b 8 9
[2 rows x 3 columns]
You also don't need to create the empty DataFrame.
The best way is to use DataFrame to construct a DF from a sequence of Series, rather than using concat:
import pandas as pd
s1 = pd.Series(['a',5,6])
s2 = pd.Series(['b',8,9])
pd.DataFrame([s1, s2])
Output:
In [4]: pd.DataFrame([s1, s2])
Out[4]:
0 1 2
0 a 5 6
1 b 8 9
A method of accomplishing the same objective as appending a Series to a DataFrame
is to just convert the data to an array of lists and append the array(s) to the DataFrame.
data as an array of lists
def get_example(idx):
list1 = (idx+1,idx+2 ,chr(idx + 97))
data = [list1]
return(data)
df1 = pd.DataFrame()
for idx in range(4):
data = get_example(idx)
df1= df1.append(data, ignore_index = True)

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