blocks|key|115171|text|首先创建布尔掩码，使用floor+by+expected，使用duplicated只保留最后一个值。然后使用mask为所有不带last的值创建NaNs。这些值被shifted和NaNs替换为正向填充。最后用Value列的第一个值替换第一个组：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|115172|first+=+df.iloc[0,+df.columns.get_loc('Value')]
m1+=+df['TimeStamp'].dt.floor('1Min').duplicated(keep='last')
df['1min']+=+df['Value'].mask(m1).shift().ffill().fillna(first)
m2+=+df['TimeStamp'].dt.floor('5Min').duplicated(keep='last')
df['5min']+=+df['Value'].mask(m2).shift().ffill().fillna(first)

print+(df)
+++++++++++++TimeStamp++Value++OneMinuteValue++FiveMinuteValue++1min++5min
0++2018-08-11+14:30:01++++5.0+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
1++2018-08-11+14:30:11++++4.8+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
2++2018-08-11+14:30:30++++4.9+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
3++2018-08-11+14:30:46++++5.0+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
4++2018-08-11+14:30:59++++5.1+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
5++2018-08-11+14:31:01++++5.2+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
6++2018-08-11+14:31:20++++5.3+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
7++2018-08-11+14:31:40++++5.4+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
8++2018-08-11+14:32:16++++5.5+++++++++++++5.4++++++++++++++++5+++5.4+++5.0
9++2018-08-11+14:32:28++++5.6+++++++++++++5.4++++++++++++++++5+++5.4+++5.0
10+2018-08-11+14:33:04++++5.6+++++++++++++5.6++++++++++++++++5+++5.6+++5.0
11+2018-08-11+14:33:40++++5.7+++++++++++++5.6++++++++++++++++5+++5.6+++5.0
12+2018-08-11+14:34:01++++5.7+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
13+2018-08-11+14:34:22++++5.8+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
14+2018-08-11+14:34:32++++5.8+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
15+2018-08-11+14:34:48++++5.9+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
16+2018-08-11+14:34:59++++6.0+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
17+2018-08-11+14:35:01++++6.1+++++++++++++6.0++++++++++++++++6+++6.0+++6.0
18+2018-08-11+14:35:01++++6.2+++++++++++++6.0++++++++++++++++5+++6.0+++6.0|code-block|syntax|javascript|115173|使用循环的另一种解决方案：|115174|cols+=+['1Min','5Min']
first+=+df.iloc[0,+df.columns.get_loc('Value')]

for+x+in+cols:
++++m+=+df['TimeStamp'].dt.floor(x).duplicated(keep='last')
++++df[x]+=+df['Value'].mask(m).shift().ffill().fillna(first)
print+(df)
+++++++++++++TimeStamp++Value++OneMinuteValue++FiveMinuteValue++1Min++5Min
0++2018-08-11+14:30:01++++5.0+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
1++2018-08-11+14:30:11++++4.8+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
2++2018-08-11+14:30:30++++4.9+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
3++2018-08-11+14:30:46++++5.0+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
4++2018-08-11+14:30:59++++5.1+++++++++++++5.0++++++++++++++++5+++5.0+++5.0
5++2018-08-11+14:31:01++++5.2+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
6++2018-08-11+14:31:20++++5.3+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
7++2018-08-11+14:31:40++++5.4+++++++++++++5.1++++++++++++++++5+++5.1+++5.0
8++2018-08-11+14:32:16++++5.5+++++++++++++5.4++++++++++++++++5+++5.4+++5.0
9++2018-08-11+14:32:28++++5.6+++++++++++++5.4++++++++++++++++5+++5.4+++5.0
10+2018-08-11+14:33:04++++5.6+++++++++++++5.6++++++++++++++++5+++5.6+++5.0
11+2018-08-11+14:33:40++++5.7+++++++++++++5.6++++++++++++++++5+++5.6+++5.0
12+2018-08-11+14:34:01++++5.7+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
13+2018-08-11+14:34:22++++5.8+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
14+2018-08-11+14:34:32++++5.8+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
15+2018-08-11+14:34:48++++5.9+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
16+2018-08-11+14:34:59++++6.0+++++++++++++5.7++++++++++++++++5+++5.7+++5.0
17+2018-08-11+14:35:01++++6.1+++++++++++++6.0++++++++++++++++6+++6.0+++6.0
18+2018-08-11+14:35:01++++6.2+++++++++++++6.0++++++++++++++++6+++6.0+++6.0|115175|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.floor.html|1|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.duplicated.html|2|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mask.html|3|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.shift.html^0|B|5|V|A|1I|4|1Z|3|28|5|2G|3|2V|5|B|5|0|V|A|1|1I|4|2|28|5|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@$9|1H|A|1I|1|1J]|$9|1K|A|1L|1|1M]|$9|1N|A|1O|1|1P]|$9|1Q|A|1R|1|1S]]|E|$]]|$1|F|3|G|5|H|7|1T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|1V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|1W|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]|Y|$5|R|S|T|E|$U|Z]]|10|$5|R|S|T|E|$U|11]]]]

First create boolean mask with <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.dt.floor.html" rel="nofollow noreferrer"><code>floor</code></a> by expected time with <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.duplicated.html" rel="nofollow noreferrer"><code>duplicated</code></a> for keep only last value. Then use <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.mask.html" rel="nofollow noreferrer"><code>mask</code></a> for create <code>NaN</code>s for all values without last. These values are <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.shift.html" rel="nofollow noreferrer"><code>shift</code></a>ed and <code>NaN</code>s are replace by forward filling. Last replace first group by first value of <code>Value</code> column:

<pre><code>first = df.iloc[0, df.columns.get_loc('Value')]
m1 = df['TimeStamp'].dt.floor('1Min').duplicated(keep='last')
df['1min'] = df['Value'].mask(m1).shift().ffill().fillna(first)
m2 = df['TimeStamp'].dt.floor('5Min').duplicated(keep='last')
df['5min'] = df['Value'].mask(m2).shift().ffill().fillna(first)

print (df)
 TimeStamp Value OneMinuteValue FiveMinuteValue 1min 5min
0 2018-08-11 14:30:01 5.0 5.0 5 5.0 5.0
1 2018-08-11 14:30:11 4.8 5.0 5 5.0 5.0
2 2018-08-11 14:30:30 4.9 5.0 5 5.0 5.0
3 2018-08-11 14:30:46 5.0 5.0 5 5.0 5.0
4 2018-08-11 14:30:59 5.1 5.0 5 5.0 5.0
5 2018-08-11 14:31:01 5.2 5.1 5 5.1 5.0
6 2018-08-11 14:31:20 5.3 5.1 5 5.1 5.0
7 2018-08-11 14:31:40 5.4 5.1 5 5.1 5.0
8 2018-08-11 14:32:16 5.5 5.4 5 5.4 5.0
9 2018-08-11 14:32:28 5.6 5.4 5 5.4 5.0
10 2018-08-11 14:33:04 5.6 5.6 5 5.6 5.0
11 2018-08-11 14:33:40 5.7 5.6 5 5.6 5.0
12 2018-08-11 14:34:01 5.7 5.7 5 5.7 5.0
13 2018-08-11 14:34:22 5.8 5.7 5 5.7 5.0
14 2018-08-11 14:34:32 5.8 5.7 5 5.7 5.0
15 2018-08-11 14:34:48 5.9 5.7 5 5.7 5.0
16 2018-08-11 14:34:59 6.0 5.7 5 5.7 5.0
17 2018-08-11 14:35:01 6.1 6.0 6 6.0 6.0
18 2018-08-11 14:35:01 6.2 6.0 5 6.0 6.0
</code></pre>

Another solution with loop:

<pre><code>cols = ['1Min','5Min']
first = df.iloc[0, df.columns.get_loc('Value')]

for x in cols:
 m = df['TimeStamp'].dt.floor(x).duplicated(keep='last')
 df[x] = df['Value'].mask(m).shift().ffill().fillna(first)
print (df)
 TimeStamp Value OneMinuteValue FiveMinuteValue 1Min 5Min
0 2018-08-11 14:30:01 5.0 5.0 5 5.0 5.0
1 2018-08-11 14:30:11 4.8 5.0 5 5.0 5.0
2 2018-08-11 14:30:30 4.9 5.0 5 5.0 5.0
3 2018-08-11 14:30:46 5.0 5.0 5 5.0 5.0
4 2018-08-11 14:30:59 5.1 5.0 5 5.0 5.0
5 2018-08-11 14:31:01 5.2 5.1 5 5.1 5.0
6 2018-08-11 14:31:20 5.3 5.1 5 5.1 5.0
7 2018-08-11 14:31:40 5.4 5.1 5 5.1 5.0
8 2018-08-11 14:32:16 5.5 5.4 5 5.4 5.0
9 2018-08-11 14:32:28 5.6 5.4 5 5.4 5.0
10 2018-08-11 14:33:04 5.6 5.6 5 5.6 5.0
11 2018-08-11 14:33:40 5.7 5.6 5 5.6 5.0
12 2018-08-11 14:34:01 5.7 5.7 5 5.7 5.0
13 2018-08-11 14:34:22 5.8 5.7 5 5.7 5.0
14 2018-08-11 14:34:32 5.8 5.7 5 5.7 5.0
15 2018-08-11 14:34:48 5.9 5.7 5 5.7 5.0
16 2018-08-11 14:34:59 6.0 5.7 5 5.7 5.0
17 2018-08-11 14:35:01 6.1 6.0 6 6.0 6.0
18 2018-08-11 14:35:01 6.2 6.0 6 6.0 6.0
</code></pre>

My data frame has two columns: TimeStamp and Value (decimal).
I want to add a 3 new columns into the dataframe:
oneMinuteValue
fiveMinuteValue
fifteenMinuteValue

The data is in sorted order of ascending timestamp. From the first timestamp I want to go through every minute from the first timestamp until the last and execute the following:

for each row that has data from this minute, set the oneMinuteValue as either:
a) the final value from the previous minute; or
b) the current Value if this timestamp happens to sit directly on the minute to the second

I want the same for the 5 and 15 minute columns, but an example with just 5 minutes will be enough I think

An example (left off the fifteen minute one):

<pre><code>TimeStamp Value OneMinuteValue FiveMinuteValue
2018-08-11 14:30:01 5 5 5
2018-08-11 14:30:11 4.8 5 5
2018-08-11 14:30:30 4.9 5 5
2018-08-11 14:30:46 5 5 5
2018-08-11 14:30:59 5.1 5 5
2018-08-11 14:31:01 5.2 5.1 5
2018-08-11 14:31:20 5.3 5.1 5
2018-08-11 14:31:40 5.4 5.1 5
2018-08-11 14:32:16 5.5 5.4 5
2018-08-11 14:32:28 5.6 5.4 5
2018-08-11 14:33:04 5.6 5.6 5
2018-08-11 14:33:40 5.7 5.6 5
2018-08-11 14:34:01 5.7 5.7 5
2018-08-11 14:34:22 5.8 5.7 5
2018-08-11 14:34:32 5.8 5.7 5
2018-08-11 14:34:48 5.9 5.7 5
2018-08-11 14:34:59 6.0 5.7 5
2018-08-11 14:35:01 6.1 6.0 6
2018-08-11 14:35:21 6.2 6.0 6
</code></pre>

How to correctly adjust a pandas dataframe row by row

我的数据框有两列: TimeStamp和Value (十进制)。我想在数据框中添加3个新列: oneMinuteValue fiveMinuteValue fifteenMinuteValue数据按时间戳升序排列。从第一个时间戳开始，我要遍历从第一个时间戳到最后一个时间戳的每一分钟，并执行以下命令：对于包含这一分钟的数...

问如何逐行正确调整pandas数据帧
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