Get First Row Value of a Given Column

Get first row value of a given column

To select the ith row, use iloc:

In [31]: df_test.iloc[0]
Out[31]: 
ATime     1.2
X         2.0
Y        15.0
Z         2.0
Btime     1.2
C        12.0
D        25.0
E        12.0
Name: 0, dtype: float64

To select the ith value in the Btime column you could use:

In [30]: df_test['Btime'].iloc[0]
Out[30]: 1.2

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There is a difference between df_test['Btime'].iloc[0] (recommended) and df_test.iloc[0]['Btime']:

DataFrames store data in column-based blocks (where each block has a single
dtype). If you select by column first, a view can be returned (which is
quicker than returning a copy) and the original dtype is preserved. In contrast,
if you select by row first, and if the DataFrame has columns of different
dtypes, then Pandas copies the data into a new Series of object dtype. So
selecting columns is a bit faster than selecting rows. Thus, although
df_test.iloc[0]['Btime'] works, df_test['Btime'].iloc[0] is a little bit
more efficient.

There is a big difference between the two when it comes to assignment.
df_test['Btime'].iloc[0] = x affects df_test, but df_test.iloc[0]['Btime']
may not. See below for an explanation of why. Because a subtle difference in
the order of indexing makes a big difference in behavior, it is better to use single indexing assignment:

df.iloc[0, df.columns.get_loc('Btime')] = x

---

df.iloc[0, df.columns.get_loc('Btime')] = x (recommended):

The recommended way to assign new values to a
DataFrame is to avoid chained indexing, and instead use the method shown by
andrew,

df.loc[df.index[n], 'Btime'] = x

or

df.iloc[n, df.columns.get_loc('Btime')] = x

The latter method is a bit faster, because df.loc has to convert the row and column labels to
positional indices, so there is a little less conversion necessary if you use
df.iloc instead.

---

df['Btime'].iloc[0] = x works, but is not recommended:

Although this works, it is taking advantage of the way DataFrames are currently implemented. There is no guarantee that Pandas has to work this way in the future. In particular, it is taking advantage of the fact that (currently) df['Btime'] always returns a
view (not a copy) so df['Btime'].iloc[n] = x can be used to assign a new value
at the nth location of the Btime column of df.

Since Pandas makes no explicit guarantees about when indexers return a view versus a copy, assignments that use chained indexing generally always raise a SettingWithCopyWarning even though in this case the assignment succeeds in modifying df:

In [22]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])
In [24]: df['bar'] = 100
In [25]: df['bar'].iloc[0] = 99
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy
  self._setitem_with_indexer(indexer, value)

In [26]: df
Out[26]: 
  foo  bar
0   A   99  <-- assignment succeeded
2   B  100
1   C  100

---

df.iloc[0]['Btime'] = x does not work:

In contrast, assignment with df.iloc[0]['bar'] = 123 does not work because df.iloc[0] is returning a copy:

In [66]: df.iloc[0]['bar'] = 123
/home/unutbu/data/binky/bin/ipython:1: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame

See the caveats in the documentation: http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing-view-versus-copy

In [67]: df
Out[67]: 
  foo  bar
0   A   99  <-- assignment failed
2   B  100
1   C  100

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Warning: I had previously suggested df_test.ix[i, 'Btime']. But this is not guaranteed to give you the ith value since ix tries to index by label before trying to index by position. So if the DataFrame has an integer index which is not in sorted order starting at 0, then using ix[i] will return the row labeled i rather than the ith row. For example,

In [1]: df = pd.DataFrame({'foo':list('ABC')}, index=[0,2,1])

In [2]: df
Out[2]: 
  foo
0   A
2   B
1   C

In [4]: df.ix[1, 'foo']
Out[4]: 'C'

pandas extract first row column value equal to 1 for each group

First filter only rows with label=1 and then remove duplicates per id by DataFrame.drop_duplicates:

df1 = df[df['label'].eq(1)].drop_duplicates('id')

Get first row of dataframe in Python Pandas based on criteria

This tutorial is a very good one for pandas slicing. Make sure you check it out. Onto some snippets... To slice a dataframe with a condition, you use this format:

>>> df[condition]

This will return a slice of your dataframe which you can index using iloc. Here are your examples:

1. Get first row where A > 3 (returns row 2)

   >>> df[df.A > 3].iloc[0]
   A    4
   B    6
   C    3
   Name: 2, dtype: int64
   

If what you actually want is the row number, rather than using iloc, it would be df[df.A > 3].index[0].

2. Get first row where A > 4 AND B > 3:

   >>> df[(df.A > 4) & (df.B > 3)].iloc[0]
   A    5
   B    4
   C    5
   Name: 4, dtype: int64
   

3. Get first row where A > 3 AND (B > 3 OR C > 2) (returns row 2)

   >>> df[(df.A > 3) & ((df.B > 3) | (df.C > 2))].iloc[0]
   A    4
   B    6
   C    3
   Name: 2, dtype: int64
   

Now, with your last case we can write a function that handles the default case of returning the descending-sorted frame:

>>> def series_or_default(X, condition, default_col, ascending=False):
...     sliced = X[condition]
...     if sliced.shape[0] == 0:
...         return X.sort_values(default_col, ascending=ascending).iloc[0]
...     return sliced.iloc[0]
>>> 
>>> series_or_default(df, df.A > 6, 'A')
A    5
B    4
C    5
Name: 4, dtype: int64

As expected, it returns row 4.

Get the first row of each group of unique values in another column

Use groupby + first:

firsts = df.groupby('col_B', as_index=False).first()

Output:

>>> firsts
  col_B  col_A
0     x      1
1    xx      2
2     y      4

If the order of the columns is important:

firsts = df.loc[df.groupby('col_B', as_index=False).first().index]

Output:

>>> firsts
   col_A col_B
0      1     x
1      2    xx
2      3    xx

Find next first row meeting a condition after a specific row in pandas

You can use:

g = df['first'].ne(df['first'].shift()).cumsum().loc[~df['first']]
# or
# g = df['first'].cumsum()[~df['first']]

out = df[df['second']].groupby(g).head(1)

Output:

   first  second
1  False    True
4  False    True

Intermediate grouper g:

1    2
3    4
4    4
5    4
7    6
Name: first, dtype: int64

What is the correct way to get the first row of a dataframe?

To get the first and last element of the column, your option is already the most efficient/correct way. If you're interested in this topic, I can recommend you to read this other Stackoverflow answer: https://stackoverflow.com/a/25254087/8294752

To get the first row, I personally prefer to use DataFrame.head(1), therefore for your code something like this:

df_first_row = sub_df.head(1)

I didn't look into how the head() method is defined in Pandas and its performance implications, but in my opinion it improves readability and reduces some potential confusion with indexes.

In other examples you might also find sub_df.iloc[0], but this option will return a pandas.Series which has as indexes the DataFrame column names.
sub_df.head(1) will return a 1-row DataFrame instead, which is the same result as sub_df.iloc[0:1,:]

Filter dataframe rows based on return value of foo() applied to first column

TLDR

mask = df.apply(lambda row: foo(row['Path']), axis=1)
res: pd.DataFrame = df[mask]

Solution

To filter the rows of a DataFrame according to the return value of foo(str: str) -> bool applied to the values contained in column Path of each row the solution is to generate a mask with pandas.DataFrame.apply().

How does a mask work?

The mask works as follow: given a dataframe df: pd.DataFrame and a mask: pd.Series<bool> accessing with square brackets df[mask] will result in a new DataFrame with only the rows corresponnding to a True value of the mask series.

How to get the mask

Since df.apply(fuction, axis, ...) takes as input a function one would be tempted to pass foo() as argument of the apply() but this is wrong.
The function argumennt of apply() must be a function taking as argument a pd.Series and not a string therefore the correct way to get the mask is the following, where axis = 1 indicates that we're applyinng the lambda to get the boolean value to every row of the dataframe rather than to every column.

mask = df.apply(lambda row: foo(row['Path']), axis=1)

How to select the first row of a specific column in a list in R

I don't think you need a list or loop. Here's my suggestion, based on my understanding of your issue:

First, define your intervals based on breaks using cut():

library(dplyr)

# Define breaks for intervals ("groups")
breaks <- seq(0, 0.28, by = 0.02)

# Group data using breaks
dat$interval <- cut(dat$Time, breaks = breaks, right = FALSE, labels = FALSE)

Now use dplyr to manipulate your data based on your condition: "find the highest velocity value in each interval of time BUT only IF it is greater than the value in the previous time interval."

dat2 <- dat %>%
  # For each group, as defined by an interval
  group_by(interval) %>%
  # Get the maximum velocity value
  mutate(interval_max = max(velocity)) %>%
  # Now ungroup and iterate across rows
  ungroup %>%
  # If the previous value is greater than the current value,
  # keep the previous value, if not, keep the current value
  mutate(interval_max_cond = ifelse(
                                 lag(interval_max) > interval_max,
                                 lag(interval_max),
                                 interval_max))

Let me know if you have any questions!

data:

dat <- structure(list(Time = c(0, 0.0099998, 0.0200003, 0.0300001, 0.0399999, 
0.0499997, 0.0600002, 0.07, 0.0799998, 0.0900003, 0.1000001, 
0.1099999, 0.1199997, 0.1300002, 0.14, 0.1499998, 0.1600003, 
0.1700001, 0.1799999, 0.1899997, 0.2000002, 0.21, 0.2199998, 
0.2300003, 0.2400001, 0.2499999, 0.2599997), velocity = c(0.3444447, 
0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 
0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 
0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 0.3444447, 
0.441667, 0.441667, 0.441667, 0.441667, 0.441667, 0.6222227, 
0.6222227, 0.6222227)), class = "data.frame", row.names = c(NA, 
-27L))

Pandas, get first and last column index for row value

You can first transpose DataFrame by DataFrame.T and then aggregate minimal and maximal index with convert values to strings by Series.dt.strftime, last convert to dictionaries by DataFrame.to_dict.

For get consecutive groups is compared shifted values with Series.cumsum.

df1 = df.T.reset_index()
L = [df1.groupby(df1[x].ne(df1[x].shift()).cumsum())
        .agg(value=(x, 'first'),
             start=('index', 'min'),
             end=('index', 'max'))
        .assign(start=lambda x: x['start'].dt.strftime('%Y-%m-%d'),
                end=lambda x: x['end'].dt.strftime('%Y-%m-%d'))
        .to_dict(orient='records') for x in df1.columns.drop('index')]
print (L)
[[{'value': 0, 'start': '2022-05-21', 'end': '2022-05-31'}, 
  {'value': 2, 'start': '2022-06-01', 'end': '2022-06-19'}, 
  {'value': 5, 'start': '2022-06-20', 'end': '2022-06-30'}],
 [{'value': 5, 'start': '2022-05-21', 'end': '2022-05-31'}, 
  {'value': 2, 'start': '2022-06-01', 'end': '2022-06-19'},
  {'value': 1, 'start': '2022-06-20', 'end': '2022-06-30'}],
 [{'value': 5, 'start': '2022-05-21', 'end': '2022-05-31'}, 
  {'value': 2, 'start': '2022-06-01', 'end': '2022-06-19'}, 
  {'value': 5, 'start': '2022-06-20', 'end': '2022-06-30'}]]

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