Reputation: 3507
I would to change the value of certain DataFrame values only if a certain condition is met an n
number of consecutive times.
Example:
df = pd.DataFrame(np.random.randn(15, 3))
df.iloc[4:8,0]=40
df.iloc[12,0]=-40
df.iloc[10:12,1]=-40
Which gives me this DF:
0 1 2
0 1.238892 0.802318 -0.013856
1 -1.136326 -0.527263 -0.260975
2 1.118771 0.031517 0.527350
3 1.629482 -0.158941 -1.045209
4 40.000000 0.598657 -1.268399
5 40.000000 0.442297 -0.016363
6 40.000000 -0.316817 1.744822
7 40.000000 0.193083 0.914172
8 0.322756 -0.680682 0.888702
9 -1.204531 -0.240042 1.416020
10 -1.337494 -40.000000 -1.195780
11 -0.703669 -40.000000 0.657519
12 -40.000000 -0.288235 -0.840145
13 -1.084869 -0.298030 -1.592004
14 -0.617568 -1.046210 -0.531523
Now, if I do
a=df.copy()
a[ abs(a) > abs(a.std()) ] = float('nan')
I get
0 1 2
0 1.238892 0.802318 -0.013856
1 -1.136326 -0.527263 -0.260975
2 1.118771 0.031517 0.527350
3 1.629482 -0.158941 NaN
4 NaN 0.598657 NaN
5 NaN 0.442297 -0.016363
6 NaN -0.316817 NaN
7 NaN 0.193083 0.914172
8 0.322756 -0.680682 0.888702
9 -1.204531 -0.240042 NaN
10 -1.337494 NaN NaN
11 -0.703669 NaN 0.657519
12 NaN -0.288235 -0.840145
13 -1.084869 -0.298030 NaN
14 -0.617568 -1.046210 -0.531523
which is fair. However, I would like only to replace the values with NaN
if these conditions were met by a maximum of 2 consecutive entries (so I can interpolate later). For example, I wanted the result to be
0 1 2
0 1.238892 0.802318 -0.013856
1 -1.136326 -0.527263 -0.260975
2 1.118771 0.031517 0.527350
3 1.629482 -0.158941 NaN
4 40.000000 0.598657 NaN
5 40.000000 0.442297 -0.016363
6 40.000000 -0.316817 NaN
7 40.000000 0.193083 0.914172
8 0.322756 -0.680682 0.888702
9 -1.204531 -0.240042 NaN
10 -1.337494 NaN NaN
11 -0.703669 NaN 0.657519
12 NaN -0.288235 -0.840145
13 -1.084869 -0.298030 NaN
14 -0.617568 -1.046210 -0.531523
Apparently there's no ready-to-use method to do this. The solution I found that closest resembles my problem was this one, but I couldn't make it work for me.
Any ideas?
Upvotes: 2
Views: 1631
Reputation: 52236
See below - the tricky part is (cond[c] != cond[c].shift(1)).cumsum()
which breaks the data into contiguous runs of the same value.
In [23]: cond = abs(df) > abs(df.std())
In [24]: for c in df.columns:
...: grouper = (cond[c] != cond[c].shift(1)).cumsum() * cond[c]
...: fill = (df.groupby(grouper)[c].transform('size') <= 2)
...: df.loc[fill, c] = np.nan
In [25]: df
Out[25]:
0 1 2
0 1.238892 0.802318 -0.013856
1 -1.136326 -0.527263 -0.260975
2 1.118771 0.031517 0.527350
3 1.629482 -0.158941 NaN
4 40.000000 0.598657 NaN
5 40.000000 0.442297 -0.016363
6 40.000000 -0.316817 NaN
7 40.000000 0.193083 0.914172
8 0.322756 -0.680682 0.888702
9 -1.204531 -0.240042 NaN
10 -1.337494 NaN NaN
11 -0.703669 NaN 0.657519
12 NaN -0.288235 -0.840145
13 -1.084869 -0.298030 NaN
14 -0.617568 -1.046210 -0.531523
To explain a bit more, cond[c]
is a boolean series indicating whether your condition is true or not.
The cond[c] != cond[c].shift(1)
compares the current row's condition to the next row's. This has the effecting of 'marking' where a run of values begins with the value True
.
The .cumsum()
converts the bools to integers and takes the cumulative sum. It may not be immediately intuitive, but this 'numbers' the groups of contiguous values. Finally the * cond[c]
reassigns all groups that didn't meet the criteria to 0 (using False == 0
)
So now you have groups of contiguous numbers that meet your condition, the next step performs a groupby
to count how many values are in each group (transform('size')
.
Finally a new bool condition is used to assign missing values to those groups with 2 or less values meeting the condition.
Upvotes: 3