I have a dataframe with a series of numbers. For example:
Index Column 1
1 10
2 12
3 24
4 NaN
5 20
6 15
7 NaN
8 NaN
9 2
I can't use bfill or ffill as the rule is dynamic, taking the value from the previous row and dividing by the number of consecutive NaN + 1. For example, rows 3 and 4 should be replaced with 12 as 24/2, rows 6, 7 and 8 should be replaced with 5. All other numbers should remain unchanged.
How should I do that?
Note: Edited the dataframe to be more general by inserting a new row between rows 4 and 5 and another row at the end.
You can do:
m = (df["Column 1"].notna()) & (
(df["Column 1"].shift(-1).isna()) | (df["Column 1"].shift().isna())
)
out = df.groupby(m.cumsum()).transform(
lambda x: x.fillna(0).mean() if x.isna().any() else x
)
print(out):
Index Column 1
0 1 10.0
1 2 12.0
2 3 12.0
3 4 12.0
4 5 20.0
5 6 5.0
6 7 5.0
7 8 5.0
8 9 2.0
Explanation and intermediate values:
Basically look for the rows where the next value is NaN or previous value is NaN but their value itself is not NaN. Those rows form the first row of such groups.
So the m in above code looks like:
0 True
1 False
2 True
3 False
4 True
5 True
6 False
7 False
8 True
now I want to form groups of rows that are ['True', <all Falses>] because those are the groups I want to take average of. For that use cumsum
If you want to take a look at those groups, you can use ngroup() after groupby on m.cumsum():
0 0
1 0
2 1
3 1
4 2
5 3
6 3
7 3
8 4
The above is only to show what are the groups.
Now for each group you can get the mean of the group if the group has any NaN value. This is accomplished by checking for NaNs using x.isna().any().
If the group has any NaN value then assign mean after filling NaN with 0 ,otherwise just keep the group as is. This is accomplished by the lambda:
lambda x: x.fillna(0).mean() if x.isna().any() else x
Why not using interpolate? There is a method=s that would probably fitsyour desire
However, if you really want to do as you described above, you can do something like this. (Note that iterating over rows in pandas is considered bad practice, but it does the job)
import pandas as pd
import numpy as np
df = pd.DataFrame([10,
12,
24,
np.NaN,
15,
np.NaN,
np.NaN])
for col in df:
for idx in df.index: # (iterating over rows is considered bad practice)
local_idx=idx
while(local_idx+1<len(df) and np.isnan(df.at[local_idx+1,col])):
local_idx+=1
if (local_idx-idx)>0:
fillvalue = df.loc[idx]/(local_idx-idx+1)
for fillidx in range(idx, local_idx+1):
df.loc[fillidx] = fillvalue
df
Output:
0
0 10.0
1 12.0
2 12.0
3 12.0
4 5.0
5 5.0
6 5.0
Related
What is the most pandastic way to create running total columns at various levels (without iterating over the rows)?
input:
import pandas as pd
import numpy as np
df = pd.DataFrame()
df['test'] = np.nan,np.nan,'X','X','X','X',np.nan,'X','X','X','X','X','X',np.nan,np.nan,'X','X'
df['desired_output_level_1'] = np.nan,np.nan,'1','1','1','1',np.nan,'2','2','2','2','2','2',np.nan,np.nan,'3','3'
df['desired_output_level_2'] = np.nan,np.nan,'1','2','3','4',np.nan,'1','2','3','4','5','6',np.nan,np.nan,'1','2'
output:
test desired_output_level_1 desired_output_level_2
0 NaN NaN NaN
1 NaN NaN NaN
2 X 1 1
3 X 1 2
4 X 1 3
5 X 1 4
6 NaN NaN NaN
7 X 2 1
8 X 2 2
9 X 2 3
10 X 2 4
11 X 2 5
12 X 2 6
13 NaN NaN NaN
14 NaN NaN NaN
15 X 3 1
16 X 3 2
The test column can only contain X's or NaNs.
The number of consecutive X's is random.
In the 'desired_output_level_1' column, trying to count up the number of series of X's.
In the 'desired_output_level_2' column, trying to find the duration of each series.
Can anyone help? Thanks in advance.
Perhaps not the most pandastic way, but seems to yield what you are after.
Three key points:
we are operating on only rows that are not NaN, so let's create a mask:
mask = df['test'].notna()
For level 1 computation, it's easy to compare when there is a change from NaN to not NaN by shifting rows by one:
df.loc[mask, "level_1"] = (df["test"].isna() & df["test"].shift(-1).notna()).cumsum()
For level 2 computation, it's a bit trickier. One way to do it is to run the computation for each level_1 group and do .transform to preserve the indexing:
df.loc[mask, "level_2"] = (
df.loc[mask, ["level_1"]]
.assign(level_2=1)
.groupby("level_1")["level_2"]
.transform("cumsum")
)
Last step (if needed) is to transform columns to strings:
df['level_1'] = df['level_1'].astype('Int64').astype('str')
df['level_2'] = df['level_2'].astype('Int64').astype('str')
I am looking how to create an if function that will select a row above another row where ColumnX is equal to the number 13.
Here is the code I have
if df.attrib.get("Column_Name") in ['13']:
I know this means that if column name "Column_Name" = 13 then ...
but I want it to be if "Column_Name" 1 row below is equal to 13 then ...
You can use simple Pandas Condition:
import pandas as pd
# i created my own dataframe for testing
df = pd.DataFrame({'numbers':[1,2,13,4,5,13,6]})
# use simple condition to get the index of the element then access the element by index
df.iloc[df[df["numbers"]==13].index-1]
+output:
numbers
1 2
4 5
You can find the columns using pandas .shift function and a .loc
>>> import pandas
>>> df = pandas.DataFrame({'condition':[1,3,4,3,2,13,3]})
>>> df
condition
0 1
1 3
2 4
3 3
4 2
5 13
6 3
>>> df["condition_rolled"] = df['condition'].shift(-1)
>>> df
condition condition_rolled
0 1 3.0
1 3 4.0
2 4 3.0 3 3 2.0
4 2 13.0
5 13 3.0
6 3 NaN
>>> df.loc[(df["condition_rolled"] == 13.0)]
condition condition_rolled
4 2 13.0
I have following df
A B
0 1 10
1 2 20
2 NaN 5
3 3 1
4 NaN 2
5 NaN 3
6 1 10
7 2 50
8 Nan 80
9 3 5
Consisting of repeating sequences from 1-3 seperated by a variable number of NaN's.I want to groupby each this sequences from 1-3 and get the minimum value of column B within these sequences.
Desired Output something like:
B_min
0 1
6 5
Many thanks beforehand
draj
Idea is first remove rows by missing values by DataFrame.dropna, then use GroupBy.cummin by helper Series created by compare A for equal by Series.eq and Series.cumsum, last data cleaning to one column DataFrame:
df = (df.dropna(subset=['A'])
.groupby(df['A'].eq(1).cumsum())['B']
.min()
.reset_index(drop=True)
.to_frame(name='B_min'))
print (df)
B_min
0 1
1 5
All you need to df.groupby() and apply min(). Is this what you are expecting?
df.groupby('A')['B'].min()
Output:
A
1 10
2 20
3 1
Nan 80
If you don't want the NaNs in your group you can drop them using df.dropna()
df.dropna().groupby('A')['B'].min()
Assume that we have the following pandas dataframe:
df = pd.DataFrame({'x':[0,0,1,0,0,0,0],'y':[1,1,1,1,1,1,0],'z':[0,1,1,1,0,0,1]})
x y z
0 0 1 0
1 0 1 1
2 1 1 1
3 0 1 1
4 0 1 0
5 0 1 0
6 0 0 1
All dataframe is filled either by 1 or 0. Looking at each column separately, if current row value is different than previous value I need to count number of previous consecutive values:
x y z
0
1 1
2 2
3 1
4 3
5
6 6 2
I tried to write a lambda function and apply it to entire dataframe, but I failed. Any idea?
Let's try this:
def f(col):
x = (col != col.shift().bfill())
s = x.cumsum()
return s.groupby(s).transform('count').shift().where(x)
df.apply(f).fillna('')
Output:
x y z
0
1 1
2 2
3 1
4 3
5
6 6 2
Details:
Use apply, to apply a custom function on each column of the dataframe.
Find the difference spots in the column then use cumsum to create groups of consecutive values, then groupby and transform to create a count for each record, then mask the values in the column using where for the difference spots.
You can try the following, where you identify the "runs" first, get the "runs" lengths. You will only entry at where it switches, so it is the lengths of the runs except the last one.
import pandas as pd
import numpy as np
def func(x,missing=np.NaN):
runs = np.cumsum(np.append(0,np.diff(x)!=0))
switches = np.where(np.diff(x!=0))[0] + 1
out = np.repeat(missing,len(x))
out[switches] = np.bincount(runs)[:-1]
# thanks to Scott see comments below
##out[switches] = pd.value_counts(runs,sort=False)[:-1]
return(out)
df.apply(func)
x y z
0 NaN NaN NaN
1 NaN NaN 1.0
2 2.0 NaN NaN
3 1.0 NaN NaN
4 NaN NaN 3.0
5 NaN NaN NaN
6 NaN 6.0 2.0
It might be faster with a good implementation of run length encoding.. but I am not too familiar with it in python..
I have a Data Frame with some values.
Suppose that does are the values of some stores, and does stores can fulfill some conditions and give them more than one 'state', but other stores can only fulfill one condition and be assigned with only one 'state'.
For example:
df = DataFrame({'one':[1,2,3,4],
'two';[5,6,7,8],
'three':[9,10,11,12]}
and these are my conditions:
df.loc[(df.one >= 1) & (df.two <= 7),'State'] = 1
df.loc[(df.one == 1) & (df.two <= 11),'State'] = 2
Three rows satisfy the first condition, but only one row satisfy the second condition.
The row that satisfy the two conditions should have in the column 'State', the state 1 and 2.
The obvious problem is that when the first condition gets assign the DataFrame looks like this:
one two three State
0 1 5 9 1.0
1 2 6 10 1.0
2 3 7 11 1.0
3 4 8 12 NaN
and when the second condition gets assign the Data Frame looks like this:
one two three State
0 1 5 9 2.0
1 2 6 10 1.0
2 3 7 11 1.0
3 4 8 12 NaN
and I want something like this:
one two three State
0 1 5 9 [1.0,2.0]
1 2 6 10 1.0
2 3 7 11 1.0
3 4 8 12 NaN
Here I used a list, but that is the idea.
And then If I do the store in a cell, how I call them, and how I use does cells with more than one value in the column 'State' for other conditions that depend in that column?
I appreciate it
This is a tricky question , I do not recommend you mixing the datatype within one column , as you showed 1st cell is type list , 1 and 2 are type int , then last one is NaN (considered as float), in that case why not make them all to list
s1=(df.one >= 1) & (df.two <= 7)
s2=(df.one == 1) & (df.two <= 11)
l=[[ z for z in [x,y] if z != 0]for x , y in zip(s1*1,s2*2)]
df['State']=l
df
Out[21]:
one two three State
0 1 5 9 [1, 2]
1 2 6 10 [1]
2 3 7 11 [1]
3 4 8 12 []