I have a series tt=pd.Series([-1,5,4,0,-7,-9]) .Now i want to sort 'tt'.
the positive values sort in assending order and negative values sort in descending order.Positive values is in front of negative values.
I want to get the following result.
4,5,0,-1,-7,-9
Is there a good way to get the result?
You want to sort on tt <= 0 first. Notice this is True for negatives and zero, False for positives. Sorting on this puts positives first. Then sort on tt.abs(). This puts the smallest sized numbers first.
df = pd.concat([tt, tt.abs(), tt.le(0)], axis=1)
df.sort_values([2, 1])[0]
2 4
1 5
3 0
0 -1
4 -7
5 -9
Name: 0, dtype: int64
This is a bit too extended but it gets you your desired output:
import pandas as pd
tt=pd.Series([-1,5,4,0,-7,-9])
pd.concat((tt[tt > 0].sort_values(ascending=True), tt[tt <= 0].sort_values(ascending=False)))
Out[1]:
0 4
1 5
2 0
3 -1
4 -7
5 -9
Hope this helps.
Related
I have a dataframe with returns from various investments in %. sort.values does not correctly order my returns. For example I just want to simply see the TEST column returns sorted by lowest to highest or vice versa. Please look at the test output, it is not correct.
df.sort_values('TEST')
gives me an output of returns that are NOT sorted correctly. Sort values code not in correct order
Also I am having an issue where it sorts positive numbers lowest to highest, then half way down starts again for negative numbers lowest to highest.
I just want it to look like following:
-3%
-1%
-0.5%
1%
2%
5%
Go for numpy.lexsort and boolean indexing :
import numpy as np
arr = np.array([float(x.rstrip("%")) for x in df["TEST"]])
idx = np.lexsort((arr,))
df = df.iloc[idx]
Output :
print(df)
TEST
0 -3%
1 -1%
2 -0.5%
3 1%
4 2%
5 5%
Input used :
df = pd.DataFrame({"TEST": ["1%", "-3%","-0.5%", "-1%", "5%", "2%"]})
TEST
0 1%
1 -3%
2 -0.5%
3 -1%
4 5%
5 2%
The issue is that the lexicographic order of string is different from the natural order (1->10->2 vs 1->2->10).
One option using the key parameter of sort_values:
df.sort_values('TEST', key=lambda s: pd.to_numeric(s.str.extract(r'(-?\d+\.?\d*)', expand=False)))
Or:
df.sort_values('TEST', key=lambda s: pd.to_numeric(s.str.rstrip('%')))
Output:
TEST
1 -3%
3 -1%
2 -0.5%
0 1%
5 2%
4 5%
I have a df containing sub-trajectories (segments) of users, with mode of travel indicated by 0,1,2... which looks like this:
df = pd.read_csv('sample.csv')
df
id lat lon mode
0 5138001 41.144540 -8.562926 0
1 5138001 41.144538 -8.562917 0
2 5138001 41.143689 -8.563012 0
3 5138003 43.131562 -8.601273 1
4 5138003 43.132107 -8.598124 1
5 5145001 37.092095 -8.205070 0
6 5145001 37.092180 -8.204872 0
7 5145015 39.289341 -8.023454 2
8 5145015 39.197432 -8.532761 2
9 5145015 39.198361 -8.375641 2
In the above sample, id is for the segments but a full trajectory maybe covered by different modes (i.e. contains multiple segments).
So the first 4-digits of id is the unique trajectories, and the last 3-digits, unique segment with that trajectory.
I know that I can count the number of unique segments in the dfusing:
df.groupby('id').['mode'].nunique()
How do I then count the number of unique trajectories 5138, 5145, ...?
Use indexing for get first 4 values with str, if necessary first convert values to strings by Series.astype:
df = df.groupby(df['id'].astype(str).str[:4])['mode'].nunique().reset_index(name='count')
print (df)
id count
0 5138 2
1 5145 2
If need processing values after first 4 ids:
s = df['id'].astype(str)
df = s.str[4:].groupby(s.str[:4]).nunique().reset_index(name='count')
print (df)
id count
0 5138 2
1 5145 2
Another idea is use lambda function:
df.groupby(df['id'].apply(lambda x: str(x)[:4]))['mode'].nunique()
I have a dataframe that contains information that is linked by an ID column. The rows are sequential with the odd rows containing a "start-point" and the even rows containing an "end" point. My goal is to collapse the data from these into a single row with columns for "start" and "end" following each other. The rows do have a "packet ID" that would link them if the sequential nature of the dataframe is not consistent.
example:
df:
0 1 2 3 4 5
0 hs6 106956570 106956648 ID_A1 60 -
1 hs1 153649721 153649769 ID_A1 60 -
2 hs1 865130744 865130819 ID_A2 0 -
3 hs7 21882206 21882237 ID_A2 0 -
4 hs1 74230744 74230819 ID_A3 0 +
5 hs8 92041314 92041508 ID_A3 0 +
The resulting dataframe that I am trying to achieve is:
new_df
0 1 2 3 4 5
0 hs6 106956570 106956648 hs1 153649721 153649769
1 hs1 865130744 865130819 hs7 21882206 21882237
2 hs1 74230744 74230819 hs8 92041314 92041508
with each row containing the information on both the start and the end-point.
I have tried to pass the IDs in to an array and use a for loop to pull the information out of the original dataframe into a new dataframe but this has not worked. I was looking at the melt documentation which would suggest that pd.melt(df, id_vars=[3], value_vars=[0,1,2]) may work but I cannot see how to get the corresponding row in to positions new_df[3,4,5].
I think that it may be something really simple that I am missing but any suggestions would be appreciated.
You can try this:
df_out = df.set_index([df.index%2, df.index//2])[df.columns[:3]]\
.unstack(0).sort_index(level=1, axis=1)
df_out.columns = np.arange(len(df_out.columns))
df_out
Output:
0 1 2 3 4 5
0 hs6 106956570 106956648 hs1 153649721 153649769
1 hs1 865130744 865130819 hs7 21882206 21882237
2 hs1 74230744 74230819 hs8 92041314 92041508
I have a dataframe with ordered times (in seconds) and a column that is either 0 or 1:
time bit
index
0 0.24 0
1 0.245 0
2 0.47 1
3 0.471 1
4 0.479 0
5 0.58 1
... ... ...
I want to select those rows where the time difference is, let's say <0.01 s. But only those differences between rows with bit 1 and bit 0. So in the above example I would only select row 3 and 4 (or any one of them). I thought that I would calculate the diff() of the time column. But I need to somehow select on the 0/1 bit.
Coming from the future to answer this one. You can apply a function to the dataframe that finds the indices of the rows that adhere to the condition and returns the row pairs accordingly:
def filter_(x, threshold = 0.01):
indices = df.index[(df.time.diff() < threshold) & (df.bit.diff().abs() == 1)]
mask = indices | indices - 1
return x[mask]
print(df.apply(filter_, args = (0.01,)))
Output:
time bit
3 0.471 1
4 0.479 0
From my "Id" Column I want to remove the one and zero's from the left.
That is
1000003 becomes 3
1000005 becomes 5
1000011 becomes 11 and so on
Ignore -1, 10 and 1000000, they will be handled as special cases. but from the remaining rows I want to remove the "1" followed by zeros.
Well you can use modulus to get the end of the numbers (they will be the remainder). So just exclude the rows with ids of [-1,10,1000000] and then compute the modulus of 1000000:
print df
Id
0 -1
1 10
2 1000000
3 1000003
4 1000005
5 1000007
6 1000009
7 1000011
keep = df.Id.isin([-1,10,1000000])
df.Id[~keep] = df.Id[~keep] % 1000000
print df
Id
0 -1
1 10
2 1000000
3 3
4 5
5 7
6 9
7 11
Edit: Here is a fully vectorized string slice version as an alternative (Like Alex' method but takes advantage of pandas' vectorized string methods):
keep = df.Id.isin([-1,10,1000000])
df.Id[~keep] = df.Id[~keep].astype(str).str[1:].astype(int)
print df
Id
0 -1
1 10
2 1000000
3 3
4 5
5 7
6 9
7 11
Here is another way you could try to do it:
def f(x):
"""convert the value to a string, then select only the characters
after the first one in the string, which is 1. For example,
100005 would be 00005 and I believe it's returning 00005.0 from
dataframe, which is why the float() is there. Then just convert
it to an int, and you'll have 5, etc.
"""
return int(float(str(x)[1:]))
# apply the function "f" to the dataframe and pass in the column 'Id'
df.apply(lambda row: f(row['Id']), axis=1)
I get that this question is satisfactory answered. But for future visitors, what I like about alex' answer is that it does not depend on there to be exactly four zeros. The accepted answer will fail if you sometimes have 10005, sometimes 1000005 and whatever.
However, to add something more to the way we think about it. If you know it's always going to be 10000, you can do
# backup all values
foo = df.id
#now, some will be negative or zero
df.id = df.id - 10000
#back in those that are negative or zero (here, first three rows)
df.if[df.if <= 0] = foo[df.id <= 0]
It gives you the same as Karl's answer, but I typically prefer these kind of methods for their readability.