Python 如何计算熊猫数据框中连续行之间的差异？
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How to calculate differences between consecutive rows in pandas data frame?
提问by scrollex
I've got a data frame, df, with three columns: count_a, count_band date; the counts are floats, and the dates are consecutive days in 2015.
我有一个df包含三列的数据框count_a,count_b和date; 计数是浮点数，日期是 2015 年的连续天数。
I'm trying to figure out the difference between each day's counts in both the count_aand count_bcolumns —?meaning, I'm trying to calculate the difference between each row and the preceding row for both of those columns. I've set the date as the index, but am having trouble figuring out how to do this; there were a couple of hints about using pd.Seriesand pd.DataFrame.diffbut I haven't had any luck finding an applicable answer or set of instructions.
我正在尝试计算count_a和count_b列中每天计数之间的差异- 意思是，我正在尝试计算这两列的每一行与前一行之间的差异。我已将日期设置为索引，但无法弄清楚如何执行此操作；有一些关于使用的提示pd.Series，pd.DataFrame.diff但我没有找到适用的答案或一组说明。
I'm a bit stuck, and would appreciate some guidance here.
我有点卡住了，希望得到一些指导。
Here's what my data frame looks like:
这是我的数据框的样子：
df=pd.Dataframe({'count_a': {Timestamp('2015-01-01 00:00:00'): 34175.0,
Timestamp('2015-01-02 00:00:00'): 72640.0,
Timestamp('2015-01-03 00:00:00'): 109354.0,
Timestamp('2015-01-04 00:00:00'): 144491.0,
Timestamp('2015-01-05 00:00:00'): 180355.0,
Timestamp('2015-01-06 00:00:00'): 214615.0,
Timestamp('2015-01-07 00:00:00'): 250096.0,
Timestamp('2015-01-08 00:00:00'): 287880.0,
Timestamp('2015-01-09 00:00:00'): 332528.0,
Timestamp('2015-01-10 00:00:00'): 381460.0,
Timestamp('2015-01-11 00:00:00'): 422981.0,
Timestamp('2015-01-12 00:00:00'): 463539.0,
Timestamp('2015-01-13 00:00:00'): 505395.0,
Timestamp('2015-01-14 00:00:00'): 549027.0,
Timestamp('2015-01-15 00:00:00'): 595377.0,
Timestamp('2015-01-16 00:00:00'): 649043.0,
Timestamp('2015-01-17 00:00:00'): 707727.0,
Timestamp('2015-01-18 00:00:00'): 761287.0,
Timestamp('2015-01-19 00:00:00'): 814372.0,
Timestamp('2015-01-20 00:00:00'): 867096.0,
Timestamp('2015-01-21 00:00:00'): 920838.0,
Timestamp('2015-01-22 00:00:00'): 983405.0,
Timestamp('2015-01-23 00:00:00'): 1067243.0,
Timestamp('2015-01-24 00:00:00'): 1164421.0,
Timestamp('2015-01-25 00:00:00'): 1252178.0,
Timestamp('2015-01-26 00:00:00'): 1341484.0,
Timestamp('2015-01-27 00:00:00'): 1427600.0,
Timestamp('2015-01-28 00:00:00'): 1511549.0,
Timestamp('2015-01-29 00:00:00'): 1594846.0,
Timestamp('2015-01-30 00:00:00'): 1694226.0,
Timestamp('2015-01-31 00:00:00'): 1806727.0,
Timestamp('2015-02-01 00:00:00'): 1899880.0,
Timestamp('2015-02-02 00:00:00'): 1987978.0,
Timestamp('2015-02-03 00:00:00'): 2080338.0,
Timestamp('2015-02-04 00:00:00'): 2175775.0,
Timestamp('2015-02-05 00:00:00'): 2279525.0,
Timestamp('2015-02-06 00:00:00'): 2403306.0,
Timestamp('2015-02-07 00:00:00'): 2545696.0,
Timestamp('2015-02-08 00:00:00'): 2672464.0,
Timestamp('2015-02-09 00:00:00'): 2794788.0},
'count_b': {Timestamp('2015-01-01 00:00:00'): nan,
Timestamp('2015-01-02 00:00:00'): nan,
Timestamp('2015-01-03 00:00:00'): nan,
Timestamp('2015-01-04 00:00:00'): nan,
Timestamp('2015-01-05 00:00:00'): nan,
Timestamp('2015-01-06 00:00:00'): nan,
Timestamp('2015-01-07 00:00:00'): nan,
Timestamp('2015-01-08 00:00:00'): nan,
Timestamp('2015-01-09 00:00:00'): nan,
Timestamp('2015-01-10 00:00:00'): nan,
Timestamp('2015-01-11 00:00:00'): nan,
Timestamp('2015-01-12 00:00:00'): nan,
Timestamp('2015-01-13 00:00:00'): nan,
Timestamp('2015-01-14 00:00:00'): nan,
Timestamp('2015-01-15 00:00:00'): nan,
Timestamp('2015-01-16 00:00:00'): nan,
Timestamp('2015-01-17 00:00:00'): nan,
Timestamp('2015-01-18 00:00:00'): nan,
Timestamp('2015-01-19 00:00:00'): nan,
Timestamp('2015-01-20 00:00:00'): nan,
Timestamp('2015-01-21 00:00:00'): nan,
Timestamp('2015-01-22 00:00:00'): nan,
Timestamp('2015-01-23 00:00:00'): nan,
Timestamp('2015-01-24 00:00:00'): 71.0,
Timestamp('2015-01-25 00:00:00'): 150.0,
Timestamp('2015-01-26 00:00:00'): 236.0,
Timestamp('2015-01-27 00:00:00'): 345.0,
Timestamp('2015-01-28 00:00:00'): 1239.0,
Timestamp('2015-01-29 00:00:00'): 2228.0,
Timestamp('2015-01-30 00:00:00'): 7094.0,
Timestamp('2015-01-31 00:00:00'): 16593.0,
Timestamp('2015-02-01 00:00:00'): 27190.0,
Timestamp('2015-02-02 00:00:00'): 37519.0,
Timestamp('2015-02-03 00:00:00'): 49003.0,
Timestamp('2015-02-04 00:00:00'): 63323.0,
Timestamp('2015-02-05 00:00:00'): 79846.0,
Timestamp('2015-02-06 00:00:00'): 101568.0,
Timestamp('2015-02-07 00:00:00'): 127120.0,
Timestamp('2015-02-08 00:00:00'): 149955.0,
Timestamp('2015-02-09 00:00:00'): 171440.0}})
采纳答案by Mike Müller
diffshould give the desired result:
diff应该给出想要的结果：
>>> df.diff()
count_a count_b
2015-01-01 NaN NaN
2015-01-02 38465 NaN
2015-01-03 36714 NaN
2015-01-04 35137 NaN
2015-01-05 35864 NaN
....
2015-02-07 142390 25552
2015-02-08 126768 22835
2015-02-09 122324 21485
回答by David Wolever
You can using the .rolling_apply(…)method:
您可以使用以下.rolling_apply(…)方法：
diffs_a = pd.rolling_apply(df['count_a'], 2, lambda x: x[0] - x[1])
Alternatively, if it's easier, you can operate on the arrays directly:
或者，如果更简单，您可以直接对数组进行操作：
count_a_vals = df['count_a'].values
diffs_a = count_a_vals[:-1] - count_a_vals[1:]
