I want to divide a dataframe by a number:
df = df/10
Is there a way to do this in a method chain?
# idea:
df = df.filter(['a','b']).query("a>100").assign(**divide by 10)
We can use DataFrame.div here:
df = df[['a','b']].query("a>100").div(10)
a b
0 40.0 0.7
1 50.0 0.8
5 70.0 0.3
Use DataFrame.pipe with lambda function for use some function for all data of DataFrame:
df = pd.DataFrame({
'a':[400,500,40,50,5,700],
'b':[7,8,9,4,2,3],
'c':[1,3,5,7,1,0],
'd':[5,3,6,9,2,4]
})
df = df.filter(['a','b']).query("a>100").pipe(lambda x: x / 10)
print (df)
a b
0 40.0 0.7
1 50.0 0.8
5 70.0 0.3
Here if use apply all columns are divided separately:
df = df.filter(['a','b']).query("a>100").apply(lambda x: x / 10)
You can see difference with print:
df1 = df.filter(['a','b']).query("a>100").pipe(lambda x: print (x))
a b
0 400 7
1 500 8
5 700 3
df2 = df.filter(['a','b']).query("a>100").apply(lambda x: print (x))
0 400
1 500
5 700
Name: a, dtype: int64
0 7
1 8
5 3
Name: b, dtype: int64
Related
I would like to take a dataframe and concatenate consecutive rows for comparison.
e.g.
Take
xyt = pd.DataFrame(np.concatenate((np.random.randn(3,2), np.arange(3).reshape((3, 1))), axis=1), columns=['x','y','t'])
Which looks something like:
x y t
0 1.237007 -1.035837 0.0
1 -1.782458 1.042942 1.0
2 0.063130 0.355014 2.0
And make:
a b
x y t x y t
0 1.237007 -1.035837 0.0 -1.782458 1.042942 1.0
1 -1.782458 1.042942 1.0 0.063130 0.355014 2.0
The best I could come up with was:
pd.DataFrame(
[np.append(x,y) for (x, y) in zip(xyt.values, xyt[1:].values)],
columns=pd.MultiIndex.from_product([('a', 'b'), xyt.columns]))
Is there a better way?
Let's try concat on axis=1 with the shifted frame:
import pandas as pd
xyt = pd.DataFrame({'x': {0: 1.237007, 1: -1.782458, 2: 0.06313},
'y': {0: -1.035837, 1: 1.042942, 2: 0.355014},
't': {0: 0.0, 1: 1.0, 2: 2.0}})
merged = pd.concat((xyt, xyt.shift(-1)), axis=1, keys=('a', 'b')).iloc[:-1]
print(merged)
merged:
a b
x y t x y t
0 1.237007 -1.035837 0.0 -1.782458 1.042942 1.0
1 -1.782458 1.042942 1.0 0.063130 0.355014 2.0
You can use pd.concat:
# Generate random data
n = 10
x, y = np.random.randn(2, n)
t = np.arange(n)
xyt = pd.DataFrame({
'x': x, 'y': y, 't': t
})
# The call
pd.concat([xyt, xyt.shift(-1)], axis=1, keys=['a','b'])
# Result
a b
x y t x y t
0 1.180544 1.707380 0 -0.227370 0.734225 1.0
1 -0.227370 0.734225 1 0.271997 -1.039424 2.0
2 0.271997 -1.039424 2 -0.729960 -1.081224 3.0
3 -0.729960 -1.081224 3 0.185301 0.530126 4.0
4 0.185301 0.530126 4 -0.175333 -0.126157 5.0
5 -0.175333 -0.126157 5 -0.634870 0.068683 6.0
6 -0.634870 0.068683 6 0.350867 0.361564 7.0
7 0.350867 0.361564 7 0.090678 -0.269504 8.0
8 0.090678 -0.269504 8 0.177076 -0.976640 9.0
9 0.177076 -0.976640 9 NaN NaN NaN
I have a pandas data-frame with a column with float numbers. I tried to split each item in a column by dot '.'. Then I want to add first items to second items. I don't know why this sample code is not working.
data=
0 28.47000
1 28.45000
2 28.16000
3 28.29000
4 28.38000
5 28.49000
6 28.21000
7 29.03000
8 29.11000
9 28.11000
new_array = []
df = list(data)
for i in np.arange(len(data)):
df1 = df[i].split('.')
df2 = df1[0]+df[1]/60
new_array=np.append(new_array,df2)
Use numpy.modf with DataFrame constructor:
arr = np.modf(data.values)
df = pd.DataFrame({'a':data, 'b':arr[1] + arr[0] / 60})
print (df)
a b
0 28.47 28.007833
1 28.45 28.007500
2 28.16 28.002667
3 28.29 28.004833
4 28.38 28.006333
5 28.49 28.008167
6 28.21 28.003500
7 29.03 29.000500
8 29.11 29.001833
9 28.11 28.001833
Detail:
arr = np.modf(data.values)
print(arr)
(array([ 0.47, 0.45, 0.16, 0.29, 0.38, 0.49, 0.21, 0.03, 0.11, 0.11]),
array([ 28., 28., 28., 28., 28., 28., 28., 29., 29., 28.]))
print(arr[0] / 60)
[ 0.00783333 0.0075 0.00266667 0.00483333 0.00633333 0.00816667
0.0035 0.0005 0.00183333 0.00183333]
EDIT:
df = pd.DataFrame({'a':data, 'b':arr[1] + arr[0]*5/3 })
print (df)
a b
0 28.47 28.783333
1 28.45 28.750000
2 28.16 28.266667
3 28.29 28.483333
4 28.38 28.633333
5 28.49 28.816667
6 28.21 28.350000
7 29.03 29.050000
8 29.11 29.183333
9 28.11 28.183333
Your data types are floats, not strings, and so cannot be .split() (this is a string method). Instead you can look to use math.modf to 'split' a float into fractional and decimal parts
https://docs.python.org/3.6/library/math.html
import math
def process(x:float, divisor:int=60) -> float:
"""
Convert a float to its constituent parts. Divide the fractional part by the divisor, and then recombine creating a 'scaled fractional' part,
"""
b, a = math.modf(x)
c = a + b/divisor
return c
df['data'].apply(process)
Out[17]:
0 28.007833
1 28.007500
2 28.002667
3 28.004833
4 28.006333
5 28.008167
6 28.003500
7 29.000500
8 29.001833
9 28.001833
Name: data=, dtype: float64
Your other option is to convert them to strings, split, convert to ints and floats again, do some maths and then combine the floats. I'd rather keep the object as it is personally.
I have this pandas df:
value
index1 index2 index3
1 1 1 10.0
2 -0.5
3 0.0
2 2 1 3.0
2 0.0
3 0.0
3 1 0.0
2 -5.0
3 6.0
I would like to get the 'value' of a specific combination of index, using a dict.
Usually, I use, for example:
df = df.iloc[df.index.isin([2],level='index1')]
df = df.iloc[df.index.isin([3],level='index2')]
df = df.iloc[df.index.isin([2],level='index3')]
value = df.values[0][0]
Now, I would like to get my value = -5 in a shorter way using this dictionary:
d = {'index1':2,'index2':3,'index3':2}
And also, if I use:
d = {'index1':2,'index2':3}
I would like to get the array:
[0.0, -5.0, 6.0]
Tips?
You can use SQL-like method DataFrame.query():
In [69]: df.query(' and '.join('{}=={}'.format(k,v) for k,v in d.items()))
Out[69]:
value
index1 index2 index3
2.0 3.0 2 -5.0
for another dict:
In [77]: d = {'index1':2,'index2':3}
In [78]: df.query(' and '.join('{}=={}'.format(k,v) for k,v in d.items()))
Out[78]:
value
index1 index2 index3
2.0 3.0 1 0.0
2 -5.0
3 6.0
A non-query way would be
In [64]: df.loc[np.logical_and.reduce([
df.index.get_level_values(k) == v for k, v in d.items()])]
Out[64]:
value
index1 index2 index3
2 3 2 -5.0
Suppose I have two-leveled multi-indexed dataframe
In [1]: index = pd.MultiIndex.from_tuples([(i,j) for i in range(3)
: for j in range(1+i)], names=list('ij') )
: df = pd.DataFrame(0.1*np.arange(2*len(index)).reshape(-1,2),
: columns=list('xy'), index=index )
: df
Out[1]:
x y
i j
0 0 0.0 0.1
1 0 0.2 0.3
1 0.4 0.5
2 0 0.6 0.7
1 0.8 0.9
2 1.0 1.1
And I want to run a custom function on every sub-dataframe:
In [2]: def my_aggr_func(subdf):
: return subdf['x'].mean() / subdf['y'].mean()
:
: level0 = df.index.levels[0].values
: pd.DataFrame({'mean_ratio': [my_aggr_func(df.loc[i]) for i in level0]},
: index=pd.Index(level0, name=index.names[0]) )
Out[2]:
mean_ratio
i
0 0.000000
1 0.750000
2 0.888889
Is there an elegant way to do it with df.groupby('i').agg(__something__) or something similar?
Need GroupBy.apply, which working with DataFrame:
df1 = df.groupby('i').apply(my_aggr_func).to_frame('mean_ratio')
print (df1)
mean_ratio
i
0 0.000000
1 0.750000
2 0.888889
You don't need the custom function. You can calculate the 'within group means' with agg then perform an eval to get the ratio you want.
df.groupby('i').agg('mean').eval('x / y')
i
0 0.000000
1 0.750000
2 0.888889
dtype: float64
What do the following assignments behave differently?
df.loc[rows, [col]] = ...
df.loc[rows, col] = ...
For example:
r = pd.DataFrame({"response": [1,1,1],},index = [1,2,3] )
df = pd.DataFrame({"x": [999,99,9],}, index = [3,4,5] )
df = pd.merge(df, r, how="left", left_index=True, right_index=True)
df.loc[df["response"].isnull(), "response"] = 0
print df
x response
3 999 0.0
4 99 0.0
5 9 0.0
but
df.loc[df["response"].isnull(), ["response"]] = 0
print df
x response
3 999 1.0
4 99 0.0
5 9 0.0
why should I expect the first to behave differently to the second?
df.loc[df["response"].isnull(), ["response"]]
returns a DataFrame, so if you want to assign something to it it must be aligned by both index and columns
Demo:
In [79]: df.loc[df["response"].isnull(), ["response"]] = \
pd.DataFrame([11,12], columns=['response'], index=[4,5])
In [80]: df
Out[80]:
x response
3 999 1.0
4 99 11.0
5 9 12.0
alternatively you can assign an array/matrix of the same shape:
In [83]: df.loc[df["response"].isnull(), ["response"]] = [11, 12]
In [84]: df
Out[84]:
x response
3 999 1.0
4 99 11.0
5 9 12.0
I'd also consider using fillna() method:
In [88]: df.response = df.response.fillna(0)
In [89]: df
Out[89]:
x response
3 999 1.0
4 99 0.0
5 9 0.0