I have dataframes I want to horizontally concatenate while ignoring the index.
I know that for arithmetic operations, ignoring the index can lead to a substantial speedup if you use the numpy array .values instead of the pandas Series. Is it possible to horizontally concatenate or merge pandas dataframes whilst ignoring the index? (To my dismay, ignore_index=True does something else.) And if so, does it give a speed gain?
import pandas as pd
df1 = pd.Series(range(10)).to_frame()
df2 = pd.Series(range(10), index=range(10, 20)).to_frame()
pd.concat([df1, df2], axis=1)
# 0 0
# 0 0.0 NaN
# 1 1.0 NaN
# 2 2.0 NaN
# 3 3.0 NaN
# 4 4.0 NaN
# 5 5.0 NaN
# 6 6.0 NaN
# 7 7.0 NaN
# 8 8.0 NaN
# 9 9.0 NaN
# 10 NaN 0.0
# 11 NaN 1.0
# 12 NaN 2.0
# 13 NaN 3.0
# 14 NaN 4.0
# 15 NaN 5.0
# 16 NaN 6.0
# 17 NaN 7.0
# 18 NaN 8.0
# 19 NaN 9.0
I know I can get the result I want by resetting the index of df2, but I wonder whether there is a faster (perhaps numpy method) to do this?
np.column_stack
Absolutely equivalent to EdChum's answer.
pd.DataFrame(
np.column_stack([df1,df2]),
columns=df1.columns.append(df2.columns)
)
0 0
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
Pandas Option with assign
You can do many things with the new columns.
I don't recommend this!
df1.assign(**df2.add_suffix('_').to_dict('l'))
0 0_
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
A pure numpy method would be to use np.hstack:
In[33]:
np.hstack([df1,df2])
Out[33]:
array([[0, 0],
[1, 1],
[2, 2],
[3, 3],
[4, 4],
[5, 5],
[6, 6],
[7, 7],
[8, 8],
[9, 9]], dtype=int64)
this can be easily converted to a df by passing this as the data arg to the DataFrame ctor:
In[34]:
pd.DataFrame(np.hstack([df1,df2]))
Out[34]:
0 1
0 0 0
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
with respect to whether the data is contiguous, the individual columns will be treated as separate arrays as it's a dict of Series essentially, as you're passing numpy arrays there is no allocation of memory and copying needed here for simple and homogeneous dtype so it should be fast.
Related
import pandas as pd
df = pd.DataFrame(
[
[5, 2],
[3, 5],
[5, 5],
[8, 9],
[90, 55]
],
columns = ['max_speed', 'shield']
)
df.loc[(df.max_speed > df.shield), ['stat', 'delta']] \
= 'overspeed', df['max_speed'] - df['shield']
I am setting multiple column using .loc as above, for some cases I get Not in index error!. Am I doing something wrong above?
Create list of tuples by same size like number of Trues with filtered Series after subtract with repeat scalar overspeed:
m = (df.max_speed > df.shield)
s = df['max_speed'] - df['shield']
df.loc[m, ['stat', 'delta']] = list(zip(['overspeed'] * m.sum(), s[m]))
print(df)
max_speed shield stat delta
0 5 2 overspeed 3.0
1 3 5 NaN NaN
2 5 5 NaN NaN
3 8 9 NaN NaN
4 90 55 overspeed 35.0
Another idea with helper DataFrame:
df.loc[m, ['stat', 'delta']] = pd.DataFrame({'stat':'overspeed', 'delta':s})[m]
Details:
print(list(zip(['overspeed'] * m.sum(), s[m])))
[('overspeed', 3), ('overspeed', 35)]
print (pd.DataFrame({'stat':'overspeed', 'delta':s})[m])
stat delta
0 overspeed 3
4 overspeed 35
Simpliest is assign separately:
df.loc[m, 'stat'] = 'overspeed'
df.loc[m, 'delta'] = df['max_speed'] - df['shield']
print(df)
max_speed shield stat delta
0 5 2 overspeed 3.0
1 3 5 NaN NaN
2 5 5 NaN NaN
3 8 9 NaN NaN
4 90 55 overspeed 35.0
I want to generate 2x6 dataframe which represents a Rack.Half of this dataframe are filled with storage items and the other half is with retrieval items.
I want to do is random chosing half of these 12 items and say that they are storage and others are retrieval.
How can I randomly choose?
I tried random.sample but this chooses random columns.Actually I want to choose random items individually.
Assuming this input:
0 1 2 3 4 5
0 0 1 2 3 4 5
1 6 7 8 9 10 11
You can craft a random numpy array to select/mask half of the values:
a = np.repeat([True,False], df.size//2)
np.random.shuffle(a)
a = a.reshape(df.shape)
Then select your two groups:
df.mask(a)
0 1 2 3 4 5
0 NaN NaN NaN 3.0 4 NaN
1 6.0 NaN 8.0 NaN 10 11.0
df.where(a)
0 1 2 3 4 5
0 0.0 1 2.0 NaN NaN 5.0
1 NaN 7 NaN 9.0 NaN NaN
If you simply want 6 random elements, use nummy.random.choice:
np.random.choice(df.to_numpy(). ravel(), 6, replace=False)
Example:
array([ 4, 5, 11, 7, 8, 3])
I have:
pd.DataFrame({'A':[1,2,3],'B':[np.NaN,4,5],'C':[6,7,np.NaN],'D':[8,9,np.NaN],'E':[np.NaN,11,12],'F':[13,14,15]})
A B C D E F
0 1 NaN 6.0 8 NaN 13
1 2 4.0 7.0 9 11.0 14
2 3 5.0 NaN NaN 12.0 15
and I want to calculate the number of non NaN sequences (cnt) and length of each run (runs). Ie, row 0 has a non NaN sequences of length 1,2, and 1, for a total of 3 sequences.
pd.DataFrame({'runs':[[1,2,1],[5],[2,2]],'cnt':[3,1,2]})
runs cnt
0 [1, 2, 1] 3
1 [5] 1
2 [2, 2] 2
Any suggestions
We can do stack then groupby with subgroup created by cumsum and isna
s = df.stack(dropna=False).reset_index(name='value')
out = s[s['value'].notna()].groupby([s['level_0'],s['value'].isna().cumsum()]).size().groupby(level=[0]).agg([list,len])
out
Out[269]:
list len
level_0
0 [1, 2, 1] 3
1 [6] 1
2 [2, 2] 2
Suppose I have a DataFrame with some NaNs:
>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df
0 1 2
0 1 2 3
1 4 NaN NaN
2 NaN NaN 9
What I need to do is replace every NaN with the first non-NaN value in the same column above it. It is assumed that the first row will never contain a NaN. So for the previous example the result would be
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9
I can just loop through the whole DataFrame column-by-column, element-by-element and set the values directly, but is there an easy (optimally a loop-free) way of achieving this?
You could use the fillna method on the DataFrame and specify the method as ffill (forward fill):
>>> df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
>>> df.fillna(method='ffill')
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9
This method...
propagate[s] last valid observation forward to next valid
To go the opposite way, there's also a bfill method.
This method doesn't modify the DataFrame inplace - you'll need to rebind the returned DataFrame to a variable or else specify inplace=True:
df.fillna(method='ffill', inplace=True)
The accepted answer is perfect. I had a related but slightly different situation where I had to fill in forward but only within groups. In case someone has the same need, know that fillna works on a DataFrameGroupBy object.
>>> example = pd.DataFrame({'number':[0,1,2,nan,4,nan,6,7,8,9],'name':list('aaabbbcccc')})
>>> example
name number
0 a 0.0
1 a 1.0
2 a 2.0
3 b NaN
4 b 4.0
5 b NaN
6 c 6.0
7 c 7.0
8 c 8.0
9 c 9.0
>>> example.groupby('name')['number'].fillna(method='ffill') # fill in row 5 but not row 3
0 0.0
1 1.0
2 2.0
3 NaN
4 4.0
5 4.0
6 6.0
7 7.0
8 8.0
9 9.0
Name: number, dtype: float64
You can use pandas.DataFrame.fillna with the method='ffill' option. 'ffill' stands for 'forward fill' and will propagate last valid observation forward. The alternative is 'bfill' which works the same way, but backwards.
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df = df.fillna(method='ffill')
print(df)
# 0 1 2
#0 1 2 3
#1 4 2 3
#2 4 2 9
There is also a direct synonym function for this, pandas.DataFrame.ffill, to make things simpler.
One thing that I noticed when trying this solution is that if you have N/A at the start or the end of the array, ffill and bfill don't quite work. You need both.
In [224]: df = pd.DataFrame([None, 1, 2, 3, None, 4, 5, 6, None])
In [225]: df.ffill()
Out[225]:
0
0 NaN
1 1.0
...
7 6.0
8 6.0
In [226]: df.bfill()
Out[226]:
0
0 1.0
1 1.0
...
7 6.0
8 NaN
In [227]: df.bfill().ffill()
Out[227]:
0
0 1.0
1 1.0
...
7 6.0
8 6.0
Only one column version
Fill NAN with last valid value
df[column_name].fillna(method='ffill', inplace=True)
Fill NAN with next valid value
df[column_name].fillna(method='backfill', inplace=True)
Just agreeing with ffill method, but one extra info is that you can limit the forward fill with keyword argument limit.
>>> import pandas as pd
>>> df = pd.DataFrame([[1, 2, 3], [None, None, 6], [None, None, 9]])
>>> df
0 1 2
0 1.0 2.0 3
1 NaN NaN 6
2 NaN NaN 9
>>> df[1].fillna(method='ffill', inplace=True)
>>> df
0 1 2
0 1.0 2.0 3
1 NaN 2.0 6
2 NaN 2.0 9
Now with limit keyword argument
>>> df[0].fillna(method='ffill', limit=1, inplace=True)
>>> df
0 1 2
0 1.0 2.0 3
1 1.0 2.0 6
2 NaN 2.0 9
ffill now has it's own method pd.DataFrame.ffill
df.ffill()
0 1 2
0 1.0 2.0 3.0
1 4.0 2.0 3.0
2 4.0 2.0 9.0
You can use fillna to remove or replace NaN values.
NaN Remove
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df.fillna(method='ffill')
0 1 2
0 1.0 2.0 3.0
1 4.0 2.0 3.0
2 4.0 2.0 9.0
NaN Replace
df.fillna(0) # 0 means What Value you want to replace
0 1 2
0 1.0 2.0 3.0
1 4.0 0.0 0.0
2 0.0 0.0 9.0
Reference pandas.DataFrame.fillna
There's also pandas.Interpolate, which I think gives one more control
import pandas as pd
df = pd.DataFrame([[1, 2, 3], [4, None, None], [None, None, 9]])
df=df.interpolate(method="pad",limit=None, downcast="infer") #downcast keeps dtype as int
print(df)
0 1 2
0 1 2 3
1 4 2 3
2 4 2 9
In my case, we have time series from different devices but some devices could not send any value during some period. So we should create NA values for every device and time period and after that do fillna.
df = pd.DataFrame([["device1", 1, 'first val of device1'], ["device2", 2, 'first val of device2'], ["device3", 3, 'first val of device3']])
df.pivot(index=1, columns=0, values=2).fillna(method='ffill').unstack().reset_index(name='value')
Result:
0 1 value
0 device1 1 first val of device1
1 device1 2 first val of device1
2 device1 3 first val of device1
3 device2 1 None
4 device2 2 first val of device2
5 device2 3 first val of device2
6 device3 1 None
7 device3 2 None
8 device3 3 first val of device3
Having a 4-D numpy.ndarray, e.g.
myarr = np.random.rand(10,4,3,2)
dims={'time':1:10,'sub':1:4,'cond':['A','B','C'],'measure':['meas1','meas2']}
But with possible higher dimensions. How can I create a pandas.dataframe with multiindex, just passing the dimensions as indexes, without further manual adjustments (reshaping the ndarray into 2D shape)?
I can't wrap my head around the reshaping, not even really in 3 dimensions quite yet, so I'm searching for an 'automatic' method if possible.
What would be a function to which to pass the column/row indexes and create a dataframe? Something like:
df=nd2df(myarr,dim2row=[0,1],dim2col=[2,3],rowlab=['time','sub'],collab=['cond','measure'])
And and up with something like:
meas1 meas2
A B C A B C
sub time
1 1
2
3
.
.
2 1
2
...
If it is not possible/feasible to do it automatized, an explanation that is less terse than the Multiindexing manual is appreciated.
I can't even get it right when I don't care about the order of the dimensions, e.g. I would expect this to work:
a=np.arange(24).reshape((3,2,2,2))
iterables=[[1,2,3],[1,2],['m1','m2'],['A','B']]
pd.MultiIndex.from_product(iterables, names=['time','sub','meas','cond'])
pd.DataFrame(a.reshape(2*3*1,2*2),index)
gives:
ValueError: Shape of passed values is (4, 6), indices imply (4, 24)
You're getting the error because you've reshaped the ndarray as 6x4 and applying an index intended to capture all dimensions in a single series. The following is a setup to get the pet example working:
a=np.arange(24).reshape((3,2,2,2))
iterables=[[1,2,3],[1,2],['m1','m2'],['A','B']]
index = pd.MultiIndex.from_product(iterables, names=['time','sub','meas','cond'])
pd.DataFrame(a.reshape(24, 1),index=index)
Solution
Here's a generic DataFrame creator that should get the job done:
def produce_df(rows, columns, row_names=None, column_names=None):
"""rows is a list of lists that will be used to build a MultiIndex
columns is a list of lists that will be used to build a MultiIndex"""
row_index = pd.MultiIndex.from_product(rows, names=row_names)
col_index = pd.MultiIndex.from_product(columns, names=column_names)
return pd.DataFrame(index=row_index, columns=col_index)
Demonstration
Without named index levels
produce_df([['a', 'b'], ['c', 'd']], [['1', '2'], ['3', '4']])
1 2
3 4 3 4
a c NaN NaN NaN NaN
d NaN NaN NaN NaN
b c NaN NaN NaN NaN
d NaN NaN NaN NaN
With named index levels
produce_df([['a', 'b'], ['c', 'd']], [['1', '2'], ['3', '4']],
row_names=['alpha1', 'alpha2'], column_names=['number1', 'number2'])
number1 1 2
number2 3 4 3 4
alpha1 alpha2
a c NaN NaN NaN NaN
d NaN NaN NaN NaN
b c NaN NaN NaN NaN
d NaN NaN NaN NaN
From the structure of your data,
names=['sub','time','measure','cond'] #ind1,ind2,col1,col2
labels=[[1,2,3],[1,2],['meas1','meas2'],list('ABC')]
A straightforward way to your goal:
index = pd.MultiIndex.from_product(labels,names=names)
data=arange(index.size) # or myarr.flatten()
df=pd.DataFrame(data,index=index)
df22=df.reset_index().pivot_table(values=0,index=names[:2],columns=names[2:])
"""
measure meas1 meas2
cond A B C A B C
sub time
1 1 0 1 2 3 4 5
2 6 7 8 9 10 11
2 1 12 13 14 15 16 17
2 18 19 20 21 22 23
3 1 24 25 26 27 28 29
2 30 31 32 33 34 35
"""
I still don't know how to do it directly, but here is an easy-to-follow step by step way:
# Create 4D-array
a=np.arange(24).reshape((3,2,2,2))
# Set only one row index
rowiter=[[1,2,3]]
row_ind=pd.MultiIndex.from_product(rowiter, names=[u'time'])
# put the rest of dimenstion into columns
coliter=[[1,2],['m1','m2'],['A','B']]
col_ind=pd.MultiIndex.from_product(coliter, names=[u'sub',u'meas',u'cond'])
ncols=np.prod([len(coliter[x]) for x in range(len(coliter))])
b=pd.DataFrame(a.reshape(len(rowiter[0]),ncols),index=row_ind,columns=col_ind)
print(b)
# Reshape columns to rows as pleased:
b=b.stack('sub')
# switch levels and order in rows (level goes from inner to outer):
c=b.swaplevel(0,1,axis=0).sortlevel(0,axis=0)
To check the correct assignment of dimensions:
print(a[:,0,0,0])
[ 0 8 16]
print(a[0,:,0,0])
[0 4]
print(a[0,0,:,0])
[0 2]
print(b)
meas m1 m2
cond A B A B
time sub
1 1 0 1 2 3
2 4 5 6 7
2 1 8 9 10 11
2 12 13 14 15
3 1 16 17 18 19
2 20 21 22 23
print(c)
meas m1 m2
cond A B A B
sub time
1 1 0 1 2 3
2 8 9 10 11
3 16 17 18 19
2 1 4 5 6 7
2 12 13 14 15
3 20 21 22 23