我如何在数据框中的整个列中将 df['2'] 除以 df['1'] ？

Question

I'm trying to find the change in nav in my dataframe by dividing each index by the previous index. I'm new to this and am stumped! Hope you can help. Thanks!

t1 = 0
d = []
for f in final_df['nav']:
    t = float(f)
    d.append(t / t1)
    t1 = t
print(d)

Answer 1

Use shift on the series.

d = final_df['nav'] / final_df['nav'].shift(-1)

You'll have to figure out what to do with the last item though, as there's nothing to divide by.

Answer 2

You can try using the .shift method for the denominator. This will move the values down by one.

In the example below what happens is
500 / NaN = NaN
415 / 500 = .83
293 / 415 = .71
... and so forth

df = df = pd.DataFrame({'value1':[500,415,293,126,115,140,90,190,217]})

# the first row will be NaN
# the first row of the df['value2'].shift() will be empty 
# if you have a value for the first row you can fill it with .fillna(value for first row denomenator)
df['new_value'] = df['value1'] / df['value1'].shift() # .fillna(value of first denomenator)
df

Answer 3

Try divided by shift

from pandas import pandas as pd

final_df = pd.DataFrame({"Col1": [10, 20, 15, 30, 45],
                   "Col2": [13, 23, 18, 33, 48],
                   "nav": [3, 6, 9, 27, 108]},
                  index=pd.date_range("2020-01-01", "2020-01-05"))
final_df["change"] = final_df["nav"].div(final_df["nav"].shift(1))

print(final_df)

            Col1  Col2  nav  change
2020-01-01    10    13    3     NaN
2020-01-02    20    23    6     2.0
2020-01-03    15    18    9     1.5
2020-01-04    30    33   27     3.0
2020-01-05    45    48  108     4.0

Answer 4

您不需要为此实现循环函数，使用pandas pct_change函数

df['nav'].pct_change()